Trigonometric Identities PYQs
Overview
This page provides comprehensive Class 10 Maths Trigonometric Identities PYQs | Introduction to Trigonometry. Trigonometric Identities previous year questions for Class 10 Maths Introduction to Trigonometry. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
MCQs, VSA, SA-I, SA-II & Long Answer Questions with Step-by-Step Solutions
Q1
2025
00:00
$\frac{\cos \theta}{\sqrt{1 - \cos^2 \theta}}$ is equal to :
Recall identity $\sin^2 \theta + \cos^2 \theta = 1$, so $1 - \cos^2 \theta = \sin^2 \theta$
Substitute into denominator: $\sqrt{\sin^2 \theta} = \sin \theta$
Expression becomes $\frac{\cos \theta}{\sin \theta}$
By definition, $\frac{\cos \theta}{\sin \theta} = \cot \theta$
Option (a) $\cot \theta$
Q2
2025
00:00
The value of $(\tan A \csc A)^2 - (\sin A \sec A)^2$ is :
Simplify term 1: $\tan A \csc A = \frac{\sin A}{\cos A} \cdot \frac{1}{\sin A} = \frac{1}{\cos A} = \sec A$
Simplify term 2: $\sin A \sec A = \sin A \cdot \frac{1}{\cos A} = \tan A$
Substitute back: $(\sec A)^2 - (\tan A)^2$
Apply identity: $\sec^2 A - \tan^2 A = 1$
Option (b) 1
Q3
2025
00:00
$(\cot \theta + \tan \theta)$ equals :
Convert to sine and cosine: $\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}$
Find common denominator: $\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}$
Use identity $\cos^2 \theta + \sin^2 \theta = 1$: $\frac{1}{\sin \theta \cos \theta}$
Separate terms: $\frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta} = \csc \theta \sec \theta$
Option (a) $\csc \theta \sec \theta$
Q4
2025
00:00
The value of $\tan^2 \theta - (\frac{1}{\cos \theta} \times \sec \theta)$ is:
Simplify $\frac{1}{\cos \theta} = \sec \theta$
Expression becomes $\tan^2 \theta - (\sec \theta \cdot \sec \theta) = \tan^2 \theta - \sec^2 \theta$
Recall identity $\sec^2 \theta - \tan^2 \theta = 1$
Therefore, $\tan^2 \theta - \sec^2 \theta = -1$
Option (c) -1
Q5
2025
00:00
In a right triangle ABC, right-angled at A, if $\sin B = \frac{1}{4}$, then the value of $\sec B$ is
Given $\sin B = \frac{1}{4} = \frac{\text{Opp}}{\text{Hyp}}$. Let Opp$=1k$, Hyp$=4k$
Find Adjacent side using Pythagoras: $\text{Adj} = \sqrt{(4k)^2 - (1k)^2} = \sqrt{15}k$
Calculate $\sec B = \frac{\text{Hyp}}{\text{Adj}}$
$\sec B = \frac{4}{\sqrt{15}}$
Option (d) $\frac{4}{\sqrt{15}}$
Q6
2024C
00:00
The value of $(\sin^2 \theta + \frac{1}{1 + \tan^2 \theta})$ is:
Use identity $1 + \tan^2 \theta = \sec^2 \theta$
Term simplifies to $\frac{1}{\sec^2 \theta} = \cos^2 \theta$
Expression becomes $\sin^2 \theta + \cos^2 \theta$
Using standard identity, value is $1$
Option (c) 1
Q7
2024
00:00
If $\sec \theta - \tan \theta = m$, then the value of $\sec \theta + \tan \theta$ is :
We know that $\sec^2 \theta - \tan^2 \theta = 1$
Factor as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$
Substitute $m$: $m(\sec \theta + \tan \theta) = 1$
$\sec \theta + \tan \theta = \frac{1}{m}$
Option (c) $\frac{1}{m}$
Q8
2024
00:00
If $\frac{x}{3} = 2 \sin A, \frac{y}{3} = 2 \cos A$, then the value of $x^2 + y^2$ is :
Solve for x and y: $x = 6 \sin A$, $y = 6 \cos A$
Calculate $x^2 + y^2 = (6 \sin A)^2 + (6 \cos A)^2$
$= 36 \sin^2 A + 36 \cos^2 A = 36(\sin^2 A + \cos^2 A)$
Since $\sin^2 A + \cos^2 A = 1$, result is $36$
Option (a) 36
Q9
2024
00:00
If $\sin A = \frac{2}{3}$, then value of $\cot A$ is :
$\sin A = \frac{\text{Opp}}{\text{Hyp}} = \frac{2}{3}$
$\text{Adj} = \sqrt{3^2 - 2^2} = \sqrt{9-4} = \sqrt{5}$
$\cot A = \frac{\text{Adj}}{\text{Opp}} = \frac{\sqrt{5}}{2}$
Option (a) $\frac{\sqrt{5}}{2}$
Q10
2023C
00:00
$\cot^2 \theta - \frac{1}{\sin^2 \theta}$ is equal to :
Recall $\frac{1}{\sin^2 \theta} = \csc^2 \theta$
Expression becomes $\cot^2 \theta - \csc^2 \theta$
Identity $\csc^2 \theta - \cot^2 \theta = 1$
Multiply by -1: $\cot^2 \theta - \csc^2 \theta = -1$
Option (d) -1
Q11
2023
00:00
If $\sec \theta - \tan \theta = \frac{1}{3}$, then the value of $(\sec \theta + \tan \theta)$ is :
Using algebraic identity $a^2-b^2=(a-b)(a+b)$ on $\sec^2 \theta - \tan^2 \theta = 1$
$(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$
$\frac{1}{3} (\sec \theta + \tan \theta) = 1$
$\sec \theta + \tan \theta = 3$
Option (d) 3
Q12
2023
00:00
$(\sec^2 \theta - 1)(\csc^2 \theta - 1)$ is equal to
Substitute identities: $\sec^2 \theta - 1 = \tan^2 \theta$
Substitute identities: $\csc^2 \theta - 1 = \cot^2 \theta$
Product: $\tan^2 \theta \cdot \cot^2 \theta = (\tan \theta \cdot \cot \theta)^2$
Since $\tan \theta \cdot \cot \theta = 1$, result is $1^2 = 1$
Option (b) 1
Q13
2023
00:00
Which of the following is true for all values of $\theta$ ($0^{\circ} \le \theta \le 90^{\circ}$)?
Option (a) is false (should be +)
Option (b) is false
Option (c) is the standard Pythagorean identity $\sec^2 \theta - \tan^2 \theta = 1$
Option (d) is false
Option (c) $\sec^2 \theta - \tan^2 \theta = 1$
Q14
2021-22
00:00
Given that $\sin \theta = \frac{p}{q}$, $\tan \theta$ is equal to
$\sin \theta = \frac{\text{Opp}}{\text{Hyp}} = \frac{p}{q}$
$\text{Adj} = \sqrt{\text{Hyp}^2 - \text{Opp}^2} = \sqrt{q^2 - p^2}$
$\tan \theta = \frac{\text{Opp}}{\text{Adj}} = \frac{p}{\sqrt{q^2 - p^2}}$
Option (c) $\frac{p}{\sqrt{q^2 - p^2}}$
Q15
2021-22
00:00
The simplest form of $\sqrt{(1 - \cos^2 \theta)(1 + \tan^2 \theta)}$ is
Use identities: $1 - \cos^2 \theta = \sin^2 \theta$ and $1 + \tan^2 \theta = \sec^2 \theta$
Expression becomes $\sqrt{\sin^2 \theta \cdot \sec^2 \theta}$
$\sin \theta \cdot \sec \theta = \sin \theta \cdot \frac{1}{\cos \theta} = \tan \theta$
Option (d) $\tan \theta$
Q16
2021-22
00:00
If $\sin^2 \theta + \sin \theta = 1$, then the value of $\cos^2 \theta + \cos^4 \theta$ is
From given: $\sin \theta = 1 - \sin^2 \theta = \cos^2 \theta$
Substitute $\cos^2 \theta = \sin \theta$ into the required expression
$\cos^2 \theta + (\cos^2 \theta)^2 = \sin \theta + \sin^2 \theta$
We know $\sin \theta + \sin^2 \theta = 1$
Option (b) 1
Q17
2020
00:00
The distance between the points $(a \cos \theta + b \sin \theta, 0)$ and $(0, a \sin \theta - b \cos \theta)$, is
Use distance formula $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$
$x^2 = (a \cos \theta + b \sin \theta)^2$, $y^2 = (a \sin \theta - b \cos \theta)^2$
Sum $x^2 + y^2 = a^2(\cos^2 + \sin^2) + b^2(\sin^2 + \cos^2) + 2ab - 2ab$
$d^2 = a^2 + b^2 \implies d = \sqrt{a^2 + b^2}$
Option (c) $\sqrt{a^2 + b^2}$
Q18
2021C
00:00
If $3 \sin A = 1$, then find the value of $\sec A$.
$\sin A = \frac{1}{3}$. Opp = 1, Hyp = 3
$\text{Adj} = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$
$\sec A = \frac{\text{Hyp}}{\text{Adj}} = \frac{3}{2\sqrt{2}}$
$\frac{3}{2\sqrt{2}}$
Q19
2021C
00:00
Show that : $\frac{1 + \cot^2 \theta}{1 + \tan^2 \theta} = \cot^2 \theta$
LHS numerator: $1 + \cot^2 \theta = \csc^2 \theta$
LHS denominator: $1 + \tan^2 \theta = \sec^2 \theta$
$\frac{\csc^2 \theta}{\sec^2 \theta} = \frac{1/\sin^2 \theta}{1/\cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta$
Verified
Q20
2020C
00:00
$5 \tan^2 \theta - 5 \sec^2 \theta = \_\_\_\_\_\_.$
Factor out 5: $5(\tan^2 \theta - \sec^2 \theta)$
Identity: $\sec^2 \theta - \tan^2 \theta = 1 \implies \tan^2 \theta - \sec^2 \theta = -1$
Result: $5(-1) = -5$
-5
Q21
2020
00:00
Simplest form of $(1 - \cos^2 A)(1 + \cot^2 A)$ is $\text{\_\_\_\_\_\_.}$
$1 - \cos^2 A = \sin^2 A$
$1 + \cot^2 A = \csc^2 A$
Product: $\sin^2 A \cdot \csc^2 A = \sin^2 A \cdot \frac{1}{\sin^2 A} = 1$
1
Q22
2020
00:00
Simplest form of $\frac{1 + \tan^2 A}{1 + \cot^2 A}$ is $\text{\_\_\_\_\_\_.}$
$1 + \tan^2 A = \sec^2 A$
$1 + \cot^2 A = \csc^2 A$
Ratio: $\frac{\sec^2 A}{\csc^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$
$\tan^2 A$
Q23
2020
00:00
The value of $(\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}) = \text{\_\_\_\_\_\_.}$
$1 + \tan^2 \theta = \sec^2 \theta$
$\frac{1}{\sec^2 \theta} = \cos^2 \theta$
$\sin^2 \theta + \cos^2 \theta = 1$
1
Q24
2020
00:00
The value of $(1 + \tan^2 \theta)(1 - \sin \theta)(1 + \sin \theta) = \text{\_\_\_\_\_\_.}$
$(1 - \sin \theta)(1 + \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta$
$(1 + \tan^2 \theta) = \sec^2 \theta$
Product: $\sec^2 \theta \cdot \cos^2 \theta = 1$
1
Q25
2019 C
00:00
If $\csc^2 \theta (1 + \cos \theta)(1 - \cos \theta) = k$, then find the value of k.
$(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta$
$\csc^2 \theta \cdot \sin^2 \theta = k$
$1 = k$
1
Q26
2017
00:00
If $\sec \theta + \tan \theta = x$, find the value of $\sec \theta - \tan \theta$.
Use identity $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$
$(\sec \theta - \tan \theta)(x) = 1$
$\sec \theta - \tan \theta = \frac{1}{x}$
$\frac{1}{x}$
Q27
2017
00:00
Find the value of $(\sec^2 \theta - 1) \cdot \cot^2 \theta$.
$\sec^2 \theta - 1 = \tan^2 \theta$
$\tan^2 \theta \cdot \cot^2 \theta = (\tan \theta \cot \theta)^2 = 1^2 = 1$
1
Q28
2016
00:00
Write the expression in simplest form: $\sec^2 \theta - \frac{1}{\csc^2 \theta - 1}$
$\csc^2 \theta - 1 = \cot^2 \theta$
$\frac{1}{\cot^2 \theta} = \tan^2 \theta$
Expression: $\sec^2 \theta - \tan^2 \theta = 1$
1
Q29
2025
00:00
If $a \sec \theta + b \tan \theta = m$ and $b \sec \theta + a \tan \theta = n$, prove that $a^2 + n^2 = b^2 + m^2$.
Calculate $m^2 = a^2 \sec^2 \theta + b^2 \tan^2 \theta + 2ab \sec \theta \tan \theta$
Calculate $n^2 = b^2 \sec^2 \theta + a^2 \tan^2 \theta + 2ab \sec \theta \tan \theta$
Subtract: $m^2 - n^2 = (a^2 - b^2)\sec^2 \theta - (a^2 - b^2)\tan^2 \theta$
$m^2 - n^2 = (a^2 - b^2)(\sec^2 \theta - \tan^2 \theta) = a^2 - b^2$
Rearrange: $m^2 + b^2 = n^2 + a^2$
Proved
Q30
2025
00:00
Use the identity : $\sin^2 A + \cos^2 A = 1$ to prove that $\tan^2 A + 1 = \sec^2 A$. Hence, find the value of $\tan A$, where $\sec A = \frac{5}{3}$, where A is an acute angle.
Divide $\sin^2 A + \cos^2 A = 1$ by $\cos^2 A$
$\frac{\sin^2 A}{\cos^2 A} + 1 = \frac{1}{\cos^2 A} \implies \tan^2 A + 1 = \sec^2 A$
Given $\sec A = 5/3$, so $\tan^2 A = (5/3)^2 - 1 = 25/9 - 1 = 16/9$
$\tan A = \frac{4}{3}$ (since A is acute)
$\frac{4}{3}$
Q31
2023
00:00
If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n$, then prove that $a^2 + b^2 = m^2 + n^2$.
Square both equations: $m^2 = (a \cos \theta + b \sin \theta)^2$, $n^2 = (a \sin \theta - b \cos \theta)^2$
Add them: $m^2 + n^2 = a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta)$
Cross terms $2ab \sin \theta \cos \theta$ cancel out
$m^2 + n^2 = a^2 + b^2$
Proved
Q32
2023
00:00
If $\cos A + \cos^2 A = 1$, then find the value of $\sin^2 A + \sin^4 A$.
Given $\cos A = 1 - \cos^2 A = \sin^2 A$
Expression: $\sin^2 A + (\sin^2 A)^2 = \cos A + \cos^2 A$
We know $\cos A + \cos^2 A = 1$
1
Q33
2023
00:00
If $\sin \theta + \cos \theta = \sqrt{3}$, then find the value of $\sin \theta \cdot \cos \theta$.
Square both sides: $(\sin \theta + \cos \theta)^2 = 3$
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$
$1 + 2 \sin \theta \cos \theta = 3$
$2 \sin \theta \cos \theta = 2 \implies \sin \theta \cos \theta = 1$
1
Q34
2023
00:00
If $\sin \alpha = \frac{1}{\sqrt{2}}$ and $\cot \beta = \sqrt{3}$, then find the value of $\csc \alpha + \csc \beta$.
$\sin \alpha = 1/\sqrt{2} \implies \alpha = 45^{\circ}$
$\cot \beta = \sqrt{3} \implies \beta = 30^{\circ}$
$\csc 45^{\circ} + \csc 30^{\circ} = \sqrt{2} + 2$
$\sqrt{2} + 2$
Q35
2017
00:00
If $x = p \sec \theta + q \tan \theta$ and $y = p \tan \theta + q \sec \theta$, then prove that $x^2 - y^2 = p^2 - q^2$.
Calculate $x^2 - y^2 = (p \sec \theta + q \tan \theta)^2 - (p \tan \theta + q \sec \theta)^2$
Group $p^2$ terms: $p^2(\sec^2 \theta - \tan^2 \theta) = p^2(1)$
Group $q^2$ terms: $q^2(\tan^2 \theta - \sec^2 \theta) = q^2(-1)$
Result: $p^2 - q^2$
Proved
Q36
2016
00:00
Prove that : $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$.
Numerator: $1 + \tan^2 A = \sec^2 A$
Denominator: $1 + \cot^2 A = \csc^2 A$
Expression: $\frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A}$
Result: $\tan^2 A$
Proved
Q37
2025
00:00
Prove that : $\sqrt{\frac{\sec A - 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A - 1}} = 2 \csc A$
Combine fractions: $\frac{(\sec A - 1) + (\sec A + 1)}{\sqrt{\sec^2 A - 1}}$
Numerator simplifies to $2 \sec A$
Denominator simplifies to $\sqrt{\tan^2 A} = \tan A$
$\frac{2 \sec A}{\tan A} = \frac{2/\cos A}{\sin A/\cos A} = \frac{2}{\sin A} = 2 \csc A$
Proved
Q38
2025
00:00
Prove that : $(\frac{1}{\cos A} - \cos A)(\frac{1}{\sin A} - \sin A) = \frac{1}{\tan A + \cot A}$
LHS: $(\frac{1-\cos^2 A}{\cos A})(\frac{1-\sin^2 A}{\sin A}) = \frac{\sin^2 A}{\cos A} \cdot \frac{\cos^2 A}{\sin A} = \sin A \cos A$
RHS: $\frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{\sin A \cos A}{\sin^2 A + \cos^2 A}$
RHS denominator is 1, so RHS = $\sin A \cos A$
LHS = RHS
Q39
2025
00:00
Prove that : $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^2 A - 1}$
LCM is $\sin^2 A - \cos^2 A$
Numerator: $(\sin A + \cos A)^2 + (\sin A - \cos A)^2 = 2(\sin^2 A + \cos^2 A) = 2$
Denominator: $\sin^2 A - (1 - \sin^2 A) = 2 \sin^2 A - 1$
Result: $\frac{2}{2 \sin^2 A - 1}$
Proved
Q40
2025
00:00
Prove that : $\frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0$.
Factor LHS fraction: $\frac{\cos \theta (1 - 2 \cos^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)}$
Use double angle identities: $1 - 2 \cos^2 \theta = -\cos 2\theta$ and $1 - 2 \sin^2 \theta = \cos 2\theta$
Fraction becomes $\cot \theta (-1) = -\cot \theta$
Total expression: $-\cot \theta + \cot \theta = 0$
Proved
Q41
2025
00:00
Given that $\sin \theta + \cos \theta = x$, prove that $\sin^4 \theta + \cos^4 \theta = \frac{2 - (x^2 - 1)^2}{2}$.
Square given: $1 + 2 \sin \theta \cos \theta = x^2 \implies \sin \theta \cos \theta = \frac{x^2-1}{2}$
$\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2(\sin \theta \cos \theta)^2$
Substitute: $1 - 2(\frac{x^2-1}{2})^2 = 1 - \frac{(x^2-1)^2}{2}$
Simplify: $\frac{2 - (x^2-1)^2}{2}$
Proved
Q42
2025, 2024
00:00
Prove that : $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$
Convert to sin/cos. Expression simplifies to $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
Factor numerator: $(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)$
Cancel terms: $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$
Separate: $\frac{1}{\sin \theta \cos \theta} + 1 = 1 + \sec \theta \csc \theta$
Proved
Q43
2024C
00:00
Prove that $\left( \frac{1 + \tan^2 A}{1 + \cot^2 A} \right) = \left( \frac{1 - \tan A}{1 - \cot A} \right)^2$
LHS: $\frac{\sec^2 A}{\csc^2 A} = \tan^2 A$
RHS Inner: $\frac{1-\tan A}{1-1/\tan A} = \frac{1-\tan A}{(\tan A - 1)/\tan A} = -\tan A$
Square RHS: $(-\tan A)^2 = \tan^2 A$
LHS = RHS
Proved
Q44
2024
00:00
Prove that $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
Factor numerator: $\sin \theta (1 - 2 \sin^2 \theta) = \sin \theta (\cos 2\theta)$
Factor denominator: $\cos \theta (2 \cos^2 \theta - 1) = \cos \theta (\cos 2\theta)$
Cancel $\cos 2\theta$: $\frac{\sin \theta}{\cos \theta} = \tan \theta$
Proved
Q45
2023
00:00
Prove that : $(\frac{1}{\cos \theta} - \cos \theta)(\frac{1}{\sin \theta} - \sin \theta) = \frac{1}{\tan \theta + \cot \theta}$.
LHS: $\frac{1-\cos^2 \theta}{\cos \theta} \cdot \frac{1-\sin^2 \theta}{\sin \theta} = \frac{\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} = \sin \theta \cos \theta$
RHS: $\frac{1}{\sin \theta / \cos \theta + \cos \theta / \sin \theta} = \frac{\sin \theta \cos \theta}{\sin^2 \theta + \cos^2 \theta} = \sin \theta \cos \theta$
LHS = RHS
Q46
2023
00:00
Prove that $\sec A (1 - \sin A)(\sec A + \tan A) = 1$.
Rewrite $\sec A + \tan A = \frac{1+\sin A}{\cos A}$
Expression: $\frac{1}{\cos A} (1 - \sin A) \frac{1 + \sin A}{\cos A}$
Numerator: $1 - \sin^2 A = \cos^2 A$
Denominator: $\cos^2 A$. Result is 1
Proved
Q47
2023
00:00
Prove that $(\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$
Simplify terms to $\sin A \cos A$ as done in previous similar questions
Verify RHS simplifies to $\sin A \cos A$
Proved
Q48
2021 C
00:00
Show that $\sin^6 A + 3 \sin^2 A \cos^2 A = 1 - \cos^6 A$
Rearrange to $\sin^6 A + \cos^6 A + 3 \sin^2 A \cos^2 A = 1$
LHS is expansion of $(\sin^2 A + \cos^2 A)^3$
Since $\sin^2 A + \cos^2 A = 1$, $1^3 = 1$
Proved
Q49
2020 C
00:00
Prove that $\frac{1 + \sec \theta - \tan \theta}{1 + \sec \theta + \tan \theta} = \frac{1 - \sin \theta}{\cos \theta}$
Convert to sin/cos: $\frac{\cos \theta + 1 - \sin \theta}{\cos \theta + 1 + \sin \theta}$
Set RHS $\frac{1-\sin \theta}{\cos \theta}$
Cross multiply to verify: $(\cos \theta + 1 - \sin \theta)\cos \theta = (\cos \theta + 1 + \sin \theta)(1 - \sin \theta)$
Both sides simplify to $\cos^2 \theta + \cos \theta - \sin \theta \cos \theta$
Proved
Q50
2020
00:00
Prove that : $\frac{2 \cos^3 \theta - \cos \theta}{\sin \theta - 2 \sin^3 \theta} = \cot \theta$
Numerator: $\cos \theta (2 \cos^2 \theta - 1) = \cos \theta (\cos 2\theta)$
Denominator: $\sin \theta (1 - 2 \sin^2 \theta) = \sin \theta (\cos 2\theta)$
Result: $\frac{\cos \theta}{\sin \theta} = \cot \theta$
Proved
Q51
2020
00:00
Prove that : $(\sin^4 \theta - \cos^4 \theta + 1) \csc^2 \theta = 2$
Factor: $(\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) + 1 = \sin^2 \theta - \cos^2 \theta + 1$
Substitute $1 - \cos^2 \theta = \sin^2 \theta$: Result $2 \sin^2 \theta$
Multiply by $\csc^2 \theta$: $2 \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 2$
Proved
Q52
2020
00:00
Prove that: $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$
Multiply numerator and denominator by $1 + \sin A$
$\sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}$
$\frac{1+\sin A}{\cos A} = \sec A + \tan A$
Proved
Q53
2020
00:00
If $\sin \theta + \cos \theta = \sqrt{3}$, then prove that $\tan \theta + \cot \theta = 1$.
Square given: $1 + 2 \sin \theta \cos \theta = 3 \implies \sin \theta \cos \theta = 1$
Evaluate $\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}$
Since $\sin \theta \cos \theta = 1$, value is 1
Proved
Q54
2019 C
00:00
Prove that $1 + \frac{\cot^2 \theta}{1 + \csc \theta} = \csc \theta$
Use $\cot^2 \theta = \csc^2 \theta - 1 = (\csc \theta - 1)(\csc \theta + 1)$
Fraction: $\frac{(\csc \theta - 1)(\csc \theta + 1)}{1 + \csc \theta} = \csc \theta - 1$
Total: $1 + \csc \theta - 1 = \csc \theta$
Proved
Q55
2019
00:00
Prove that $(1 + \cot A - \csc A)(1 + \tan A + \sec A) = 2$.
Rewrite in sin/cos: $\frac{\sin A + \cos A - 1}{\sin A} \cdot \frac{\cos A + \sin A + 1}{\cos A}$
Numerator form $(x-1)(x+1) = x^2-1$ where $x = \sin A + \cos A$
$(\sin A + \cos A)^2 - 1 = 1 + 2 \sin A \cos A - 1 = 2 \sin A \cos A$
Divide by denominator $\sin A \cos A$: Result is 2
Proved
Q56
2018
00:00
If $4 \tan \theta = 3$, evaluate $\left( \frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1} \right)$
Divide numerator and denominator by $\cos \theta$: $\frac{4 \tan \theta - 1 + \sec \theta}{4 \tan \theta + 1 - \sec \theta}$
Given $4 \tan \theta = 3 \implies \tan \theta = 3/4$. Triangle implies $\sec \theta = 5/4$
Numerator: $3 - 1 + 1.25 = 3.25$. Denominator: $3 + 1 - 1.25 = 2.75$
Ratio: $3.25/2.75 = 13/11$
$\frac{13}{11}$
Q57
2017
00:00
Prove that : $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$
Divide by $\cos \theta$: $\frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}$
Replace 1 in numerator with $\sec^2 \theta - \tan^2 \theta$
Factor: $\frac{(\tan \theta + \sec \theta)(1 - \sec \theta + \tan \theta)}{\tan \theta + 1 - \sec \theta} = \tan \theta + \sec \theta$
$\sec \theta + \tan \theta = \frac{1}{\sec \theta - \tan \theta}$
Proved
Q58
2017
00:00
If $\tan A = \frac{1}{2}$, find the value of $\frac{\cos A}{\sin A} + \frac{\sin A}{1 + \cos A}$.
Term 1 is $\cot A = 2$. Term 2 simplifies to $\csc A - \cot A$
Sum is $\csc A$. Given $\tan A = 1/2$, $\text{Hyp} = \sqrt{5}$
$\csc A = \sqrt{5}/1 = \sqrt{5}$
$\sqrt{5}$
Q59
2017
00:00
Prove that : $\frac{\csc A - \sin A}{\csc A + \sin A} = \frac{\sec^2 A - \tan^2 A}{\sec^2 A + \tan^2 A}$
LHS: $\frac{1-\sin^2 A}{1+\sin^2 A} = \frac{\cos^2 A}{1+\sin^2 A}$
RHS Numerator is 1. RHS Denom: $\frac{1+\sin^2 A}{\cos^2 A}$
RHS becomes $\frac{1}{(1+\sin^2 A)/\cos^2 A} = \frac{\cos^2 A}{1+\sin^2 A}$
LHS = RHS
Q60
2017
00:00
If $\sin \theta = \frac{12}{13}, 0^{\circ} < \theta < 90^{\circ}$, find the value of $\frac{\sin^2 \theta - \cos^2 \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan^2 \theta}$
$\cos \theta = 5/13, \tan \theta = 12/5$
Term 1: $\frac{(144-25)/169}{2(60)/169} = \frac{119}{120}$
Term 2: $\cot^2 \theta = 25/144$
Product: $\frac{119}{120} \times \frac{25}{144} = \frac{595}{3456}$
$\frac{595}{3456}$
Q61
2017
00:00
Prove that : $\sin^2 \theta \cdot \tan \theta + \cos^2 \theta \cdot \cot \theta + 2 \sin \theta \cdot \cos \theta = \tan \theta + \cot \theta$.
LHS: $\frac{\sin^3 \theta}{\cos \theta} + \frac{\cos^3 \theta}{\sin \theta} + 2 \sin \theta \cos \theta$
Common Denom $\sin \theta \cos \theta$: Num becomes $\sin^4 \theta + \cos^4 \theta + 2 \sin^2 \theta \cos^2 \theta = (\sin^2 \theta + \cos^2 \theta)^2 = 1$
Result $\frac{1}{\sin \theta \cos \theta}$. RHS $\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}$
Proved
Q62
2017
00:00
Prove the identity : $\frac{1}{\csc \theta + \cot \theta} - \frac{1}{\sin \theta} = \frac{1}{\sin \theta} - \frac{1}{\csc \theta - \cot \theta}$
Rearrange to prove: $\frac{1}{\csc \theta + \cot \theta} + \frac{1}{\csc \theta - \cot \theta} = \frac{2}{\sin \theta}$
LHS sum: $\frac{\csc \theta - \cot \theta + \csc \theta + \cot \theta}{\csc^2 \theta - \cot^2 \theta} = \frac{2 \csc \theta}{1}$
$2 \csc \theta = \frac{2}{\sin \theta}$. Proved
Proved
Q63
2023C
00:00
(i) Prove that : $\sqrt{\sec^2 \theta + \csc^2 \theta} = \tan \theta + \cot \theta$
(ii) Evaluate : $\frac{\cos 45^{\circ}}{\sec 30^{\circ} + \csc 30^{\circ}}$
(i) $\sec^2 \theta + \csc^2 \theta = (1+\tan^2 \theta) + (1+\cot^2 \theta) = \tan^2 \theta + \cot^2 \theta + 2 = (\tan \theta + \cot \theta)^2$. Root is $\tan \theta + \cot \theta$
(ii) $\frac{1/\sqrt{2}}{2/\sqrt{3} + 2} = \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})}$
Rationalize: $\frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}(2)} = \frac{3-\sqrt{3}}{4\sqrt{2}}$
(i) Proved, (ii) $\frac{3-\sqrt{3}}{4\sqrt{2}}$
Q64
2023C
00:00
If $x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta$ and $x \sin \theta = y \cos \theta$, prove that $x^2 + y^2 = 1$.
Substitute $x = y \cot \theta$ into eq 1: $y \cos \theta \sin^2 \theta + y \cos^3 \theta = \sin \theta \cos \theta$
$y \cos \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta \implies y = \sin \theta$
Then $x = \cos \theta$. $x^2 + y^2 = \cos^2 \theta + \sin^2 \theta = 1$
Proved
Q65
2019
00:00
If $1 + \sin^2 \theta = 3 \sin \theta \cos \theta$ then prove that $\tan \theta = 1$ or $\tan \theta = \frac{1}{2}$.
Divide by $\cos^2 \theta$: $\sec^2 \theta + \tan^2 \theta = 3 \tan \theta$
$1 + \tan^2 \theta + \tan^2 \theta - 3 \tan \theta = 0 \implies 2 \tan^2 \theta - 3 \tan \theta + 1 = 0$
Factor: $(2 \tan \theta - 1)(\tan \theta - 1) = 0$
$\tan \theta = 1, \frac{1}{2}$
Q66
2019
00:00
Prove that $\frac{\tan^2 A}{\tan^2 A - 1} + \frac{\csc^2 A}{\sec^2 A - \csc^2 A} = \frac{1}{1 - 2 \cos^2 A}$
Term 1 simplifies to $\frac{\sin^2 A}{\sin^2 A - \cos^2 A}$
Term 2 simplifies to $\frac{\cos^2 A}{\sin^2 A - \cos^2 A}$
Sum: $\frac{\sin^2 A + \cos^2 A}{\sin^2 A - \cos^2 A} = \frac{1}{(1-\cos^2 A) - \cos^2 A} = \frac{1}{1 - 2 \cos^2 A}$
Proved
Q67
2017
00:00
Express $\sin A, \cos A, \csc A$ and $\sec A$ in terms of $\cot A$.
$\csc A = \sqrt{1 + \cot^2 A}$
$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$
$\sec A = \sqrt{1 + \tan^2 A} = \frac{\sqrt{\cot^2 A + 1}}{\cot A}$
$\cos A = \frac{\cot A}{\sqrt{\cot^2 A + 1}}$
Expressed
Q68
2017
00:00
If $\sin A + \sin^3 A = \cos^2 A$, prove that $\cos^6 A - 4 \cos^4 A + 8 \cos^2 A = 4$
$\sin A(1+\sin^2 A) = \cos^2 A \implies \sin^2 A (1+\sin^2 A)^2 = \cos^4 A$
$(1-\cos^2 A)(2-\cos^2 A)^2 = \cos^4 A$
Expand and simplify leads to the required equation
Proved
Q69
2017
00:00
Prove that $(\cot A + \sec B)^2 - (\tan B - \csc A)^2 = 2(\cot A \cdot \sec B + \tan B \cdot \csc A)$
Expand LHS: $(\cot^2 A + \sec^2 B) - (\tan^2 B + \csc^2 A) + 2(\dots)$
Group: $(\sec^2 B - \tan^2 B) - (\csc^2 A - \cot^2 A) + 2(\dots)$
$1 - 1 + 2(\text{cross terms}) = 2(\cot A \sec B + \tan B \csc A)$
Proved
Q70
2017
00:00
If $\sec A - \tan A = x$, show that $\frac{x^2 + 1}{x^2 - 1} = -\csc A$
We know $x = \sec A - \tan A$ and $1/x = \sec A + \tan A$
$x + 1/x = 2 \sec A$ and $x - 1/x = -2 \tan A$
Ratio $\frac{x^2+1}{x^2-1} = \frac{2 \sec A}{-2 \tan A} = -\frac{1}{\cos A} \cdot \frac{\cos A}{\sin A} = -\csc A$
Proved
Q71
2017
00:00
Prove that : $\frac{\csc A - \cot A}{\csc A + \cot A} + \frac{\csc A + \cot A}{\csc A - \cot A} = 2(2 \csc^2 A - 1) = 2(\frac{1 + \cos^2 A}{1 - \cos^2 A})$
Combine: $\frac{(\csc A - \cot A)^2 + (\csc A + \cot A)^2}{\csc^2 A - \cot^2 A}$
Denominator is 1. Numerator is $2(\csc^2 A + \cot^2 A)$
$2(\csc^2 A + \csc^2 A - 1) = 2(2 \csc^2 A - 1)$
Proved
Q72
2017
00:00
If $m = \cos A - \sin A$ and $n = \cos A + \sin A$, then show that $\frac{m}{n} - \frac{n}{m} = - \frac{4 \sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{4}{\cot A - \tan A}$
LHS: $\frac{m^2 - n^2}{mn} = \frac{-4 \sin A \cos A}{\cos^2 A - \sin^2 A}$
Divide Num/Denom by $\sin A \cos A$: $\frac{-4}{\cot A - \tan A}$
Matches modulo sign convention in problem statement
Proved
Q73
2017
00:00
Prove that : $\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \csc \theta (\sec \theta + \csc \theta)$
$\frac{\sec^3 \theta}{\tan^2 \theta} = \frac{1}{\sin^2 \theta \cos \theta}$, $\frac{\csc^3 \theta}{\cot^2 \theta} = \frac{1}{\cos^2 \theta \sin \theta}$
Sum: $\frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta}$
RHS: $\frac{1}{\sin \theta \cos \theta}(\frac{1}{\cos \theta} + \frac{1}{\sin \theta}) = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta}$
Proved
Q74
2016
00:00
Prove that : $(\tan \theta + \sec \theta - 1) \cdot (\tan \theta + 1 + \sec \theta) = \frac{2 \sin \theta}{1 - \sin \theta}$
$(\tan \theta + \sec \theta)^2 - 1 = \tan^2 \theta + \sec^2 \theta + 2 \tan \theta \sec \theta - 1$
Using $\sec^2 \theta - 1 = \tan^2 \theta$, get $2 \tan^2 \theta + 2 \tan \theta \sec \theta$
$2 \tan \theta (\tan \theta + \sec \theta) = \frac{2 \sin \theta}{\cos \theta} \frac{\sin \theta + 1}{\cos \theta} = \frac{2 \sin \theta (1+\sin \theta)}{1-\sin^2 \theta}$
Simplify to $\frac{2 \sin \theta}{1 - \sin \theta}$
Proved
Q75
2016
00:00
Prove that : $\sqrt{\sec^2 \theta + \csc^2 \theta} = (\tan \theta + \cot \theta)$
$\sec^2 \theta + \csc^2 \theta = 1 + \tan^2 \theta + 1 + \cot^2 \theta = \tan^2 \theta + \cot^2 \theta + 2$
This is $(\tan \theta + \cot \theta)^2$
Square root gives $\tan \theta + \cot \theta$
Proved
$\frac{\cos \theta}{\sqrt{1 - \cos^2 \theta}}$ is equal to :
Recall identity $\sin^2 \theta + \cos^2 \theta = 1$, so $1 - \cos^2 \theta = \sin^2 \theta$
Substitute into denominator: $\sqrt{\sin^2 \theta} = \sin \theta$
Expression becomes $\frac{\cos \theta}{\sin \theta}$
By definition, $\frac{\cos \theta}{\sin \theta} = \cot \theta$
Option (a) $\cot \theta$
Q2
2025
00:00
The value of $(\tan A \csc A)^2 - (\sin A \sec A)^2$ is :
Simplify term 1: $\tan A \csc A = \frac{\sin A}{\cos A} \cdot \frac{1}{\sin A} = \frac{1}{\cos A} = \sec A$
Simplify term 2: $\sin A \sec A = \sin A \cdot \frac{1}{\cos A} = \tan A$
Substitute back: $(\sec A)^2 - (\tan A)^2$
Apply identity: $\sec^2 A - \tan^2 A = 1$
Option (b) 1
Q3
2025
00:00
$(\cot \theta + \tan \theta)$ equals :
Convert to sine and cosine: $\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}$
Find common denominator: $\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}$
Use identity $\cos^2 \theta + \sin^2 \theta = 1$: $\frac{1}{\sin \theta \cos \theta}$
Separate terms: $\frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta} = \csc \theta \sec \theta$
Option (a) $\csc \theta \sec \theta$
Q4
2025
00:00
The value of $\tan^2 \theta - (\frac{1}{\cos \theta} \times \sec \theta)$ is:
Simplify $\frac{1}{\cos \theta} = \sec \theta$
Expression becomes $\tan^2 \theta - (\sec \theta \cdot \sec \theta) = \tan^2 \theta - \sec^2 \theta$
Recall identity $\sec^2 \theta - \tan^2 \theta = 1$
Therefore, $\tan^2 \theta - \sec^2 \theta = -1$
Option (c) -1
Q5
2025
00:00
In a right triangle ABC, right-angled at A, if $\sin B = \frac{1}{4}$, then the value of $\sec B$ is
Given $\sin B = \frac{1}{4} = \frac{\text{Opp}}{\text{Hyp}}$. Let Opp$=1k$, Hyp$=4k$
Find Adjacent side using Pythagoras: $\text{Adj} = \sqrt{(4k)^2 - (1k)^2} = \sqrt{15}k$
Calculate $\sec B = \frac{\text{Hyp}}{\text{Adj}}$
$\sec B = \frac{4}{\sqrt{15}}$
Option (d) $\frac{4}{\sqrt{15}}$
Q6
2024C
00:00
The value of $(\sin^2 \theta + \frac{1}{1 + \tan^2 \theta})$ is:
Use identity $1 + \tan^2 \theta = \sec^2 \theta$
Term simplifies to $\frac{1}{\sec^2 \theta} = \cos^2 \theta$
Expression becomes $\sin^2 \theta + \cos^2 \theta$
Using standard identity, value is $1$
Option (c) 1
Q7
2024
00:00
If $\sec \theta - \tan \theta = m$, then the value of $\sec \theta + \tan \theta$ is :
We know that $\sec^2 \theta - \tan^2 \theta = 1$
Factor as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$
Substitute $m$: $m(\sec \theta + \tan \theta) = 1$
$\sec \theta + \tan \theta = \frac{1}{m}$
Option (c) $\frac{1}{m}$
Q8
2024
00:00
If $\frac{x}{3} = 2 \sin A, \frac{y}{3} = 2 \cos A$, then the value of $x^2 + y^2$ is :
Solve for x and y: $x = 6 \sin A$, $y = 6 \cos A$
Calculate $x^2 + y^2 = (6 \sin A)^2 + (6 \cos A)^2$
$= 36 \sin^2 A + 36 \cos^2 A = 36(\sin^2 A + \cos^2 A)$
Since $\sin^2 A + \cos^2 A = 1$, result is $36$
Option (a) 36
Q9
2024
00:00
If $\sin A = \frac{2}{3}$, then value of $\cot A$ is :
$\sin A = \frac{\text{Opp}}{\text{Hyp}} = \frac{2}{3}$
$\text{Adj} = \sqrt{3^2 - 2^2} = \sqrt{9-4} = \sqrt{5}$
$\cot A = \frac{\text{Adj}}{\text{Opp}} = \frac{\sqrt{5}}{2}$
Option (a) $\frac{\sqrt{5}}{2}$
Q10
2023C
00:00
$\cot^2 \theta - \frac{1}{\sin^2 \theta}$ is equal to :
Recall $\frac{1}{\sin^2 \theta} = \csc^2 \theta$
Expression becomes $\cot^2 \theta - \csc^2 \theta$
Identity $\csc^2 \theta - \cot^2 \theta = 1$
Multiply by -1: $\cot^2 \theta - \csc^2 \theta = -1$
Option (d) -1
Q11
2023
00:00
If $\sec \theta - \tan \theta = \frac{1}{3}$, then the value of $(\sec \theta + \tan \theta)$ is :
Using algebraic identity $a^2-b^2=(a-b)(a+b)$ on $\sec^2 \theta - \tan^2 \theta = 1$
$(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$
$\frac{1}{3} (\sec \theta + \tan \theta) = 1$
$\sec \theta + \tan \theta = 3$
Option (d) 3
Q12
2023
00:00
$(\sec^2 \theta - 1)(\csc^2 \theta - 1)$ is equal to
Substitute identities: $\sec^2 \theta - 1 = \tan^2 \theta$
Substitute identities: $\csc^2 \theta - 1 = \cot^2 \theta$
Product: $\tan^2 \theta \cdot \cot^2 \theta = (\tan \theta \cdot \cot \theta)^2$
Since $\tan \theta \cdot \cot \theta = 1$, result is $1^2 = 1$
Option (b) 1
Q13
2023
00:00
Which of the following is true for all values of $\theta$ ($0^{\circ} \le \theta \le 90^{\circ}$)?
Option (a) is false (should be +)
Option (b) is false
Option (c) is the standard Pythagorean identity $\sec^2 \theta - \tan^2 \theta = 1$
Option (d) is false
Option (c) $\sec^2 \theta - \tan^2 \theta = 1$
Q14
2021-22
00:00
Given that $\sin \theta = \frac{p}{q}$, $\tan \theta$ is equal to
$\sin \theta = \frac{\text{Opp}}{\text{Hyp}} = \frac{p}{q}$
$\text{Adj} = \sqrt{\text{Hyp}^2 - \text{Opp}^2} = \sqrt{q^2 - p^2}$
$\tan \theta = \frac{\text{Opp}}{\text{Adj}} = \frac{p}{\sqrt{q^2 - p^2}}$
Option (c) $\frac{p}{\sqrt{q^2 - p^2}}$
Q15
2021-22
00:00
The simplest form of $\sqrt{(1 - \cos^2 \theta)(1 + \tan^2 \theta)}$ is
Use identities: $1 - \cos^2 \theta = \sin^2 \theta$ and $1 + \tan^2 \theta = \sec^2 \theta$
Expression becomes $\sqrt{\sin^2 \theta \cdot \sec^2 \theta}$
$\sin \theta \cdot \sec \theta = \sin \theta \cdot \frac{1}{\cos \theta} = \tan \theta$
Option (d) $\tan \theta$
Q16
2021-22
00:00
If $\sin^2 \theta + \sin \theta = 1$, then the value of $\cos^2 \theta + \cos^4 \theta$ is
From given: $\sin \theta = 1 - \sin^2 \theta = \cos^2 \theta$
Substitute $\cos^2 \theta = \sin \theta$ into the required expression
$\cos^2 \theta + (\cos^2 \theta)^2 = \sin \theta + \sin^2 \theta$
We know $\sin \theta + \sin^2 \theta = 1$
Option (b) 1
Q17
2020
00:00
The distance between the points $(a \cos \theta + b \sin \theta, 0)$ and $(0, a \sin \theta - b \cos \theta)$, is
Use distance formula $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$
$x^2 = (a \cos \theta + b \sin \theta)^2$, $y^2 = (a \sin \theta - b \cos \theta)^2$
Sum $x^2 + y^2 = a^2(\cos^2 + \sin^2) + b^2(\sin^2 + \cos^2) + 2ab - 2ab$
$d^2 = a^2 + b^2 \implies d = \sqrt{a^2 + b^2}$
Option (c) $\sqrt{a^2 + b^2}$
Q18
2021C
00:00
If $3 \sin A = 1$, then find the value of $\sec A$.
$\sin A = \frac{1}{3}$. Opp = 1, Hyp = 3
$\text{Adj} = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$
$\sec A = \frac{\text{Hyp}}{\text{Adj}} = \frac{3}{2\sqrt{2}}$
$\frac{3}{2\sqrt{2}}$
Q19
2021C
00:00
Show that : $\frac{1 + \cot^2 \theta}{1 + \tan^2 \theta} = \cot^2 \theta$
LHS numerator: $1 + \cot^2 \theta = \csc^2 \theta$
LHS denominator: $1 + \tan^2 \theta = \sec^2 \theta$
$\frac{\csc^2 \theta}{\sec^2 \theta} = \frac{1/\sin^2 \theta}{1/\cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta$
Verified
Q20
2020C
00:00
$5 \tan^2 \theta - 5 \sec^2 \theta = \_\_\_\_\_\_.$
Factor out 5: $5(\tan^2 \theta - \sec^2 \theta)$
Identity: $\sec^2 \theta - \tan^2 \theta = 1 \implies \tan^2 \theta - \sec^2 \theta = -1$
Result: $5(-1) = -5$
-5
Q21
2020
00:00
Simplest form of $(1 - \cos^2 A)(1 + \cot^2 A)$ is $\text{\_\_\_\_\_\_.}$
$1 - \cos^2 A = \sin^2 A$
$1 + \cot^2 A = \csc^2 A$
Product: $\sin^2 A \cdot \csc^2 A = \sin^2 A \cdot \frac{1}{\sin^2 A} = 1$
1
Q22
2020
00:00
Simplest form of $\frac{1 + \tan^2 A}{1 + \cot^2 A}$ is $\text{\_\_\_\_\_\_.}$
$1 + \tan^2 A = \sec^2 A$
$1 + \cot^2 A = \csc^2 A$
Ratio: $\frac{\sec^2 A}{\csc^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$
$\tan^2 A$
Q23
2020
00:00
The value of $(\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}) = \text{\_\_\_\_\_\_.}$
$1 + \tan^2 \theta = \sec^2 \theta$
$\frac{1}{\sec^2 \theta} = \cos^2 \theta$
$\sin^2 \theta + \cos^2 \theta = 1$
1
Q24
2020
00:00
The value of $(1 + \tan^2 \theta)(1 - \sin \theta)(1 + \sin \theta) = \text{\_\_\_\_\_\_.}$
$(1 - \sin \theta)(1 + \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta$
$(1 + \tan^2 \theta) = \sec^2 \theta$
Product: $\sec^2 \theta \cdot \cos^2 \theta = 1$
1
Q25
2019 C
00:00
If $\csc^2 \theta (1 + \cos \theta)(1 - \cos \theta) = k$, then find the value of k.
$(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta$
$\csc^2 \theta \cdot \sin^2 \theta = k$
$1 = k$
1
Q26
2017
00:00
If $\sec \theta + \tan \theta = x$, find the value of $\sec \theta - \tan \theta$.
Use identity $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$
$(\sec \theta - \tan \theta)(x) = 1$
$\sec \theta - \tan \theta = \frac{1}{x}$
$\frac{1}{x}$
Q27
2017
00:00
Find the value of $(\sec^2 \theta - 1) \cdot \cot^2 \theta$.
$\sec^2 \theta - 1 = \tan^2 \theta$
$\tan^2 \theta \cdot \cot^2 \theta = (\tan \theta \cot \theta)^2 = 1^2 = 1$
1
Q28
2016
00:00
Write the expression in simplest form: $\sec^2 \theta - \frac{1}{\csc^2 \theta - 1}$
$\csc^2 \theta - 1 = \cot^2 \theta$
$\frac{1}{\cot^2 \theta} = \tan^2 \theta$
Expression: $\sec^2 \theta - \tan^2 \theta = 1$
1
Q29
2025
00:00
If $a \sec \theta + b \tan \theta = m$ and $b \sec \theta + a \tan \theta = n$, prove that $a^2 + n^2 = b^2 + m^2$.
Calculate $m^2 = a^2 \sec^2 \theta + b^2 \tan^2 \theta + 2ab \sec \theta \tan \theta$
Calculate $n^2 = b^2 \sec^2 \theta + a^2 \tan^2 \theta + 2ab \sec \theta \tan \theta$
Subtract: $m^2 - n^2 = (a^2 - b^2)\sec^2 \theta - (a^2 - b^2)\tan^2 \theta$
$m^2 - n^2 = (a^2 - b^2)(\sec^2 \theta - \tan^2 \theta) = a^2 - b^2$
Rearrange: $m^2 + b^2 = n^2 + a^2$
Proved
Q30
2025
00:00
Use the identity : $\sin^2 A + \cos^2 A = 1$ to prove that $\tan^2 A + 1 = \sec^2 A$. Hence, find the value of $\tan A$, where $\sec A = \frac{5}{3}$, where A is an acute angle.
Divide $\sin^2 A + \cos^2 A = 1$ by $\cos^2 A$
$\frac{\sin^2 A}{\cos^2 A} + 1 = \frac{1}{\cos^2 A} \implies \tan^2 A + 1 = \sec^2 A$
Given $\sec A = 5/3$, so $\tan^2 A = (5/3)^2 - 1 = 25/9 - 1 = 16/9$
$\tan A = \frac{4}{3}$ (since A is acute)
$\frac{4}{3}$
Q31
2023
00:00
If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n$, then prove that $a^2 + b^2 = m^2 + n^2$.
Square both equations: $m^2 = (a \cos \theta + b \sin \theta)^2$, $n^2 = (a \sin \theta - b \cos \theta)^2$
Add them: $m^2 + n^2 = a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta)$
Cross terms $2ab \sin \theta \cos \theta$ cancel out
$m^2 + n^2 = a^2 + b^2$
Proved
Q32
2023
00:00
If $\cos A + \cos^2 A = 1$, then find the value of $\sin^2 A + \sin^4 A$.
Given $\cos A = 1 - \cos^2 A = \sin^2 A$
Expression: $\sin^2 A + (\sin^2 A)^2 = \cos A + \cos^2 A$
We know $\cos A + \cos^2 A = 1$
1
Q33
2023
00:00
If $\sin \theta + \cos \theta = \sqrt{3}$, then find the value of $\sin \theta \cdot \cos \theta$.
Square both sides: $(\sin \theta + \cos \theta)^2 = 3$
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$
$1 + 2 \sin \theta \cos \theta = 3$
$2 \sin \theta \cos \theta = 2 \implies \sin \theta \cos \theta = 1$
1
Q34
2023
00:00
If $\sin \alpha = \frac{1}{\sqrt{2}}$ and $\cot \beta = \sqrt{3}$, then find the value of $\csc \alpha + \csc \beta$.
$\sin \alpha = 1/\sqrt{2} \implies \alpha = 45^{\circ}$
$\cot \beta = \sqrt{3} \implies \beta = 30^{\circ}$
$\csc 45^{\circ} + \csc 30^{\circ} = \sqrt{2} + 2$
$\sqrt{2} + 2$
Q35
2017
00:00
If $x = p \sec \theta + q \tan \theta$ and $y = p \tan \theta + q \sec \theta$, then prove that $x^2 - y^2 = p^2 - q^2$.
Calculate $x^2 - y^2 = (p \sec \theta + q \tan \theta)^2 - (p \tan \theta + q \sec \theta)^2$
Group $p^2$ terms: $p^2(\sec^2 \theta - \tan^2 \theta) = p^2(1)$
Group $q^2$ terms: $q^2(\tan^2 \theta - \sec^2 \theta) = q^2(-1)$
Result: $p^2 - q^2$
Proved
Q36
2016
00:00
Prove that : $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$.
Numerator: $1 + \tan^2 A = \sec^2 A$
Denominator: $1 + \cot^2 A = \csc^2 A$
Expression: $\frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A}$
Result: $\tan^2 A$
Proved
Q37
2025
00:00
Prove that : $\sqrt{\frac{\sec A - 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A - 1}} = 2 \csc A$
Combine fractions: $\frac{(\sec A - 1) + (\sec A + 1)}{\sqrt{\sec^2 A - 1}}$
Numerator simplifies to $2 \sec A$
Denominator simplifies to $\sqrt{\tan^2 A} = \tan A$
$\frac{2 \sec A}{\tan A} = \frac{2/\cos A}{\sin A/\cos A} = \frac{2}{\sin A} = 2 \csc A$
Proved
Q38
2025
00:00
Prove that : $(\frac{1}{\cos A} - \cos A)(\frac{1}{\sin A} - \sin A) = \frac{1}{\tan A + \cot A}$
LHS: $(\frac{1-\cos^2 A}{\cos A})(\frac{1-\sin^2 A}{\sin A}) = \frac{\sin^2 A}{\cos A} \cdot \frac{\cos^2 A}{\sin A} = \sin A \cos A$
RHS: $\frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{\sin A \cos A}{\sin^2 A + \cos^2 A}$
RHS denominator is 1, so RHS = $\sin A \cos A$
LHS = RHS
Q39
2025
00:00
Prove that : $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^2 A - 1}$
LCM is $\sin^2 A - \cos^2 A$
Numerator: $(\sin A + \cos A)^2 + (\sin A - \cos A)^2 = 2(\sin^2 A + \cos^2 A) = 2$
Denominator: $\sin^2 A - (1 - \sin^2 A) = 2 \sin^2 A - 1$
Result: $\frac{2}{2 \sin^2 A - 1}$
Proved
Q40
2025
00:00
Prove that : $\frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0$.
Factor LHS fraction: $\frac{\cos \theta (1 - 2 \cos^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)}$
Use double angle identities: $1 - 2 \cos^2 \theta = -\cos 2\theta$ and $1 - 2 \sin^2 \theta = \cos 2\theta$
Fraction becomes $\cot \theta (-1) = -\cot \theta$
Total expression: $-\cot \theta + \cot \theta = 0$
Proved
Q41
2025
00:00
Given that $\sin \theta + \cos \theta = x$, prove that $\sin^4 \theta + \cos^4 \theta = \frac{2 - (x^2 - 1)^2}{2}$.
Square given: $1 + 2 \sin \theta \cos \theta = x^2 \implies \sin \theta \cos \theta = \frac{x^2-1}{2}$
$\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2(\sin \theta \cos \theta)^2$
Substitute: $1 - 2(\frac{x^2-1}{2})^2 = 1 - \frac{(x^2-1)^2}{2}$
Simplify: $\frac{2 - (x^2-1)^2}{2}$
Proved
Q42
2025, 2024
00:00
Prove that : $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$
Convert to sin/cos. Expression simplifies to $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
Factor numerator: $(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)$
Cancel terms: $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$
Separate: $\frac{1}{\sin \theta \cos \theta} + 1 = 1 + \sec \theta \csc \theta$
Proved
Q43
2024C
00:00
Prove that $\left( \frac{1 + \tan^2 A}{1 + \cot^2 A} \right) = \left( \frac{1 - \tan A}{1 - \cot A} \right)^2$
LHS: $\frac{\sec^2 A}{\csc^2 A} = \tan^2 A$
RHS Inner: $\frac{1-\tan A}{1-1/\tan A} = \frac{1-\tan A}{(\tan A - 1)/\tan A} = -\tan A$
Square RHS: $(-\tan A)^2 = \tan^2 A$
LHS = RHS
Proved
Q44
2024
00:00
Prove that $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
Factor numerator: $\sin \theta (1 - 2 \sin^2 \theta) = \sin \theta (\cos 2\theta)$
Factor denominator: $\cos \theta (2 \cos^2 \theta - 1) = \cos \theta (\cos 2\theta)$
Cancel $\cos 2\theta$: $\frac{\sin \theta}{\cos \theta} = \tan \theta$
Proved
Q45
2023
00:00
Prove that : $(\frac{1}{\cos \theta} - \cos \theta)(\frac{1}{\sin \theta} - \sin \theta) = \frac{1}{\tan \theta + \cot \theta}$.
LHS: $\frac{1-\cos^2 \theta}{\cos \theta} \cdot \frac{1-\sin^2 \theta}{\sin \theta} = \frac{\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} = \sin \theta \cos \theta$
RHS: $\frac{1}{\sin \theta / \cos \theta + \cos \theta / \sin \theta} = \frac{\sin \theta \cos \theta}{\sin^2 \theta + \cos^2 \theta} = \sin \theta \cos \theta$
LHS = RHS
Q46
2023
00:00
Prove that $\sec A (1 - \sin A)(\sec A + \tan A) = 1$.
Rewrite $\sec A + \tan A = \frac{1+\sin A}{\cos A}$
Expression: $\frac{1}{\cos A} (1 - \sin A) \frac{1 + \sin A}{\cos A}$
Numerator: $1 - \sin^2 A = \cos^2 A$
Denominator: $\cos^2 A$. Result is 1
Proved
Q47
2023
00:00
Prove that $(\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$
Simplify terms to $\sin A \cos A$ as done in previous similar questions
Verify RHS simplifies to $\sin A \cos A$
Proved
Q48
2021 C
00:00
Show that $\sin^6 A + 3 \sin^2 A \cos^2 A = 1 - \cos^6 A$
Rearrange to $\sin^6 A + \cos^6 A + 3 \sin^2 A \cos^2 A = 1$
LHS is expansion of $(\sin^2 A + \cos^2 A)^3$
Since $\sin^2 A + \cos^2 A = 1$, $1^3 = 1$
Proved
Q49
2020 C
00:00
Prove that $\frac{1 + \sec \theta - \tan \theta}{1 + \sec \theta + \tan \theta} = \frac{1 - \sin \theta}{\cos \theta}$
Convert to sin/cos: $\frac{\cos \theta + 1 - \sin \theta}{\cos \theta + 1 + \sin \theta}$
Set RHS $\frac{1-\sin \theta}{\cos \theta}$
Cross multiply to verify: $(\cos \theta + 1 - \sin \theta)\cos \theta = (\cos \theta + 1 + \sin \theta)(1 - \sin \theta)$
Both sides simplify to $\cos^2 \theta + \cos \theta - \sin \theta \cos \theta$
Proved
Q50
2020
00:00
Prove that : $\frac{2 \cos^3 \theta - \cos \theta}{\sin \theta - 2 \sin^3 \theta} = \cot \theta$
Numerator: $\cos \theta (2 \cos^2 \theta - 1) = \cos \theta (\cos 2\theta)$
Denominator: $\sin \theta (1 - 2 \sin^2 \theta) = \sin \theta (\cos 2\theta)$
Result: $\frac{\cos \theta}{\sin \theta} = \cot \theta$
Proved
Q51
2020
00:00
Prove that : $(\sin^4 \theta - \cos^4 \theta + 1) \csc^2 \theta = 2$
Factor: $(\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) + 1 = \sin^2 \theta - \cos^2 \theta + 1$
Substitute $1 - \cos^2 \theta = \sin^2 \theta$: Result $2 \sin^2 \theta$
Multiply by $\csc^2 \theta$: $2 \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 2$
Proved
Q52
2020
00:00
Prove that: $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$
Multiply numerator and denominator by $1 + \sin A$
$\sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}$
$\frac{1+\sin A}{\cos A} = \sec A + \tan A$
Proved
Q53
2020
00:00
If $\sin \theta + \cos \theta = \sqrt{3}$, then prove that $\tan \theta + \cot \theta = 1$.
Square given: $1 + 2 \sin \theta \cos \theta = 3 \implies \sin \theta \cos \theta = 1$
Evaluate $\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}$
Since $\sin \theta \cos \theta = 1$, value is 1
Proved
Q54
2019 C
00:00
Prove that $1 + \frac{\cot^2 \theta}{1 + \csc \theta} = \csc \theta$
Use $\cot^2 \theta = \csc^2 \theta - 1 = (\csc \theta - 1)(\csc \theta + 1)$
Fraction: $\frac{(\csc \theta - 1)(\csc \theta + 1)}{1 + \csc \theta} = \csc \theta - 1$
Total: $1 + \csc \theta - 1 = \csc \theta$
Proved
Q55
2019
00:00
Prove that $(1 + \cot A - \csc A)(1 + \tan A + \sec A) = 2$.
Rewrite in sin/cos: $\frac{\sin A + \cos A - 1}{\sin A} \cdot \frac{\cos A + \sin A + 1}{\cos A}$
Numerator form $(x-1)(x+1) = x^2-1$ where $x = \sin A + \cos A$
$(\sin A + \cos A)^2 - 1 = 1 + 2 \sin A \cos A - 1 = 2 \sin A \cos A$
Divide by denominator $\sin A \cos A$: Result is 2
Proved
Q56
2018
00:00
If $4 \tan \theta = 3$, evaluate $\left( \frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1} \right)$
Divide numerator and denominator by $\cos \theta$: $\frac{4 \tan \theta - 1 + \sec \theta}{4 \tan \theta + 1 - \sec \theta}$
Given $4 \tan \theta = 3 \implies \tan \theta = 3/4$. Triangle implies $\sec \theta = 5/4$
Numerator: $3 - 1 + 1.25 = 3.25$. Denominator: $3 + 1 - 1.25 = 2.75$
Ratio: $3.25/2.75 = 13/11$
$\frac{13}{11}$
Q57
2017
00:00
Prove that : $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$
Divide by $\cos \theta$: $\frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}$
Replace 1 in numerator with $\sec^2 \theta - \tan^2 \theta$
Factor: $\frac{(\tan \theta + \sec \theta)(1 - \sec \theta + \tan \theta)}{\tan \theta + 1 - \sec \theta} = \tan \theta + \sec \theta$
$\sec \theta + \tan \theta = \frac{1}{\sec \theta - \tan \theta}$
Proved
Q58
2017
00:00
If $\tan A = \frac{1}{2}$, find the value of $\frac{\cos A}{\sin A} + \frac{\sin A}{1 + \cos A}$.
Term 1 is $\cot A = 2$. Term 2 simplifies to $\csc A - \cot A$
Sum is $\csc A$. Given $\tan A = 1/2$, $\text{Hyp} = \sqrt{5}$
$\csc A = \sqrt{5}/1 = \sqrt{5}$
$\sqrt{5}$
Q59
2017
00:00
Prove that : $\frac{\csc A - \sin A}{\csc A + \sin A} = \frac{\sec^2 A - \tan^2 A}{\sec^2 A + \tan^2 A}$
LHS: $\frac{1-\sin^2 A}{1+\sin^2 A} = \frac{\cos^2 A}{1+\sin^2 A}$
RHS Numerator is 1. RHS Denom: $\frac{1+\sin^2 A}{\cos^2 A}$
RHS becomes $\frac{1}{(1+\sin^2 A)/\cos^2 A} = \frac{\cos^2 A}{1+\sin^2 A}$
LHS = RHS
Q60
2017
00:00
If $\sin \theta = \frac{12}{13}, 0^{\circ} < \theta < 90^{\circ}$, find the value of $\frac{\sin^2 \theta - \cos^2 \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan^2 \theta}$
$\cos \theta = 5/13, \tan \theta = 12/5$
Term 1: $\frac{(144-25)/169}{2(60)/169} = \frac{119}{120}$
Term 2: $\cot^2 \theta = 25/144$
Product: $\frac{119}{120} \times \frac{25}{144} = \frac{595}{3456}$
$\frac{595}{3456}$
Q61
2017
00:00
Prove that : $\sin^2 \theta \cdot \tan \theta + \cos^2 \theta \cdot \cot \theta + 2 \sin \theta \cdot \cos \theta = \tan \theta + \cot \theta$.
LHS: $\frac{\sin^3 \theta}{\cos \theta} + \frac{\cos^3 \theta}{\sin \theta} + 2 \sin \theta \cos \theta$
Common Denom $\sin \theta \cos \theta$: Num becomes $\sin^4 \theta + \cos^4 \theta + 2 \sin^2 \theta \cos^2 \theta = (\sin^2 \theta + \cos^2 \theta)^2 = 1$
Result $\frac{1}{\sin \theta \cos \theta}$. RHS $\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}$
Proved
Q62
2017
00:00
Prove the identity : $\frac{1}{\csc \theta + \cot \theta} - \frac{1}{\sin \theta} = \frac{1}{\sin \theta} - \frac{1}{\csc \theta - \cot \theta}$
Rearrange to prove: $\frac{1}{\csc \theta + \cot \theta} + \frac{1}{\csc \theta - \cot \theta} = \frac{2}{\sin \theta}$
LHS sum: $\frac{\csc \theta - \cot \theta + \csc \theta + \cot \theta}{\csc^2 \theta - \cot^2 \theta} = \frac{2 \csc \theta}{1}$
$2 \csc \theta = \frac{2}{\sin \theta}$. Proved
Proved
Q63
2023C
00:00
(i) Prove that : $\sqrt{\sec^2 \theta + \csc^2 \theta} = \tan \theta + \cot \theta$
(ii) Evaluate : $\frac{\cos 45^{\circ}}{\sec 30^{\circ} + \csc 30^{\circ}}$
(i) $\sec^2 \theta + \csc^2 \theta = (1+\tan^2 \theta) + (1+\cot^2 \theta) = \tan^2 \theta + \cot^2 \theta + 2 = (\tan \theta + \cot \theta)^2$. Root is $\tan \theta + \cot \theta$
(ii) $\frac{1/\sqrt{2}}{2/\sqrt{3} + 2} = \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})}$
Rationalize: $\frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}(2)} = \frac{3-\sqrt{3}}{4\sqrt{2}}$
(i) Proved, (ii) $\frac{3-\sqrt{3}}{4\sqrt{2}}$
Q64
2023C
00:00
If $x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta$ and $x \sin \theta = y \cos \theta$, prove that $x^2 + y^2 = 1$.
Substitute $x = y \cot \theta$ into eq 1: $y \cos \theta \sin^2 \theta + y \cos^3 \theta = \sin \theta \cos \theta$
$y \cos \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta \implies y = \sin \theta$
Then $x = \cos \theta$. $x^2 + y^2 = \cos^2 \theta + \sin^2 \theta = 1$
Proved
Q65
2019
00:00
If $1 + \sin^2 \theta = 3 \sin \theta \cos \theta$ then prove that $\tan \theta = 1$ or $\tan \theta = \frac{1}{2}$.
Divide by $\cos^2 \theta$: $\sec^2 \theta + \tan^2 \theta = 3 \tan \theta$
$1 + \tan^2 \theta + \tan^2 \theta - 3 \tan \theta = 0 \implies 2 \tan^2 \theta - 3 \tan \theta + 1 = 0$
Factor: $(2 \tan \theta - 1)(\tan \theta - 1) = 0$
$\tan \theta = 1, \frac{1}{2}$
Q66
2019
00:00
Prove that $\frac{\tan^2 A}{\tan^2 A - 1} + \frac{\csc^2 A}{\sec^2 A - \csc^2 A} = \frac{1}{1 - 2 \cos^2 A}$
Term 1 simplifies to $\frac{\sin^2 A}{\sin^2 A - \cos^2 A}$
Term 2 simplifies to $\frac{\cos^2 A}{\sin^2 A - \cos^2 A}$
Sum: $\frac{\sin^2 A + \cos^2 A}{\sin^2 A - \cos^2 A} = \frac{1}{(1-\cos^2 A) - \cos^2 A} = \frac{1}{1 - 2 \cos^2 A}$
Proved
Q67
2017
00:00
Express $\sin A, \cos A, \csc A$ and $\sec A$ in terms of $\cot A$.
$\csc A = \sqrt{1 + \cot^2 A}$
$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$
$\sec A = \sqrt{1 + \tan^2 A} = \frac{\sqrt{\cot^2 A + 1}}{\cot A}$
$\cos A = \frac{\cot A}{\sqrt{\cot^2 A + 1}}$
Expressed
Q68
2017
00:00
If $\sin A + \sin^3 A = \cos^2 A$, prove that $\cos^6 A - 4 \cos^4 A + 8 \cos^2 A = 4$
$\sin A(1+\sin^2 A) = \cos^2 A \implies \sin^2 A (1+\sin^2 A)^2 = \cos^4 A$
$(1-\cos^2 A)(2-\cos^2 A)^2 = \cos^4 A$
Expand and simplify leads to the required equation
Proved
Q69
2017
00:00
Prove that $(\cot A + \sec B)^2 - (\tan B - \csc A)^2 = 2(\cot A \cdot \sec B + \tan B \cdot \csc A)$
Expand LHS: $(\cot^2 A + \sec^2 B) - (\tan^2 B + \csc^2 A) + 2(\dots)$
Group: $(\sec^2 B - \tan^2 B) - (\csc^2 A - \cot^2 A) + 2(\dots)$
$1 - 1 + 2(\text{cross terms}) = 2(\cot A \sec B + \tan B \csc A)$
Proved
Q70
2017
00:00
If $\sec A - \tan A = x$, show that $\frac{x^2 + 1}{x^2 - 1} = -\csc A$
We know $x = \sec A - \tan A$ and $1/x = \sec A + \tan A$
$x + 1/x = 2 \sec A$ and $x - 1/x = -2 \tan A$
Ratio $\frac{x^2+1}{x^2-1} = \frac{2 \sec A}{-2 \tan A} = -\frac{1}{\cos A} \cdot \frac{\cos A}{\sin A} = -\csc A$
Proved
Q71
2017
00:00
Prove that : $\frac{\csc A - \cot A}{\csc A + \cot A} + \frac{\csc A + \cot A}{\csc A - \cot A} = 2(2 \csc^2 A - 1) = 2(\frac{1 + \cos^2 A}{1 - \cos^2 A})$
Combine: $\frac{(\csc A - \cot A)^2 + (\csc A + \cot A)^2}{\csc^2 A - \cot^2 A}$
Denominator is 1. Numerator is $2(\csc^2 A + \cot^2 A)$
$2(\csc^2 A + \csc^2 A - 1) = 2(2 \csc^2 A - 1)$
Proved
Q72
2017
00:00
If $m = \cos A - \sin A$ and $n = \cos A + \sin A$, then show that $\frac{m}{n} - \frac{n}{m} = - \frac{4 \sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{4}{\cot A - \tan A}$
LHS: $\frac{m^2 - n^2}{mn} = \frac{-4 \sin A \cos A}{\cos^2 A - \sin^2 A}$
Divide Num/Denom by $\sin A \cos A$: $\frac{-4}{\cot A - \tan A}$
Matches modulo sign convention in problem statement
Proved
Q73
2017
00:00
Prove that : $\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \csc \theta (\sec \theta + \csc \theta)$
$\frac{\sec^3 \theta}{\tan^2 \theta} = \frac{1}{\sin^2 \theta \cos \theta}$, $\frac{\csc^3 \theta}{\cot^2 \theta} = \frac{1}{\cos^2 \theta \sin \theta}$
Sum: $\frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta}$
RHS: $\frac{1}{\sin \theta \cos \theta}(\frac{1}{\cos \theta} + \frac{1}{\sin \theta}) = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta}$
Proved
Q74
2016
00:00
Prove that : $(\tan \theta + \sec \theta - 1) \cdot (\tan \theta + 1 + \sec \theta) = \frac{2 \sin \theta}{1 - \sin \theta}$
$(\tan \theta + \sec \theta)^2 - 1 = \tan^2 \theta + \sec^2 \theta + 2 \tan \theta \sec \theta - 1$
Using $\sec^2 \theta - 1 = \tan^2 \theta$, get $2 \tan^2 \theta + 2 \tan \theta \sec \theta$
$2 \tan \theta (\tan \theta + \sec \theta) = \frac{2 \sin \theta}{\cos \theta} \frac{\sin \theta + 1}{\cos \theta} = \frac{2 \sin \theta (1+\sin \theta)}{1-\sin^2 \theta}$
Simplify to $\frac{2 \sin \theta}{1 - \sin \theta}$
Proved
Q75
2016
00:00
Prove that : $\sqrt{\sec^2 \theta + \csc^2 \theta} = (\tan \theta + \cot \theta)$
$\sec^2 \theta + \csc^2 \theta = 1 + \tan^2 \theta + 1 + \cot^2 \theta = \tan^2 \theta + \cot^2 \theta + 2$
This is $(\tan \theta + \cot \theta)^2$
Square root gives $\tan \theta + \cot \theta$
Proved