Values Of Trigonometric Ratios PYQs
Overview
This page provides comprehensive Class 10 Maths Values Of Trigonometric Ratios PYQs | Introduction to Trigonometry. Values Of Trigonometric Ratios 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
If $\tan 3\theta = \sqrt{3}$, then $\theta/2$ equals
Given: $\tan 3\theta = \sqrt{3}$
We know that $\tan 60^{\circ} = \sqrt{3}$
So, $3\theta = 60^{\circ} \implies \theta = 20^{\circ}$
Required value: $\theta/2 = 20^{\circ}/2 = 10^{\circ}$
Option (d) $10^{\circ}$
Q2
2025
00:00
If $\sin 4\theta = \frac{\sqrt{3}}{2}$, then $\theta/3$ equals :
Given: $\sin 4\theta = \frac{\sqrt{3}}{2}$
We know that $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$
So, $4\theta = 60^{\circ} \implies \theta = 15^{\circ}$
Required value: $\theta/3 = 15^{\circ}/3 = 5^{\circ}$
Option (d) $5^{\circ}$
Q3
2025
00:00
If $\alpha + \beta = 90^{\circ}$ and $\alpha = 2\beta$, then $\cos^2 \alpha + \sin^2 \beta$ is equal to :
Substitute $\alpha = 2\beta$ into $\alpha + \beta = 90^{\circ}$
$3\beta = 90^{\circ} \implies \beta = 30^{\circ}$
Then $\alpha = 60^{\circ}$
Expression: $\cos^2 60^{\circ} + \sin^2 30^{\circ} = (\frac{1}{2})^2 + (\frac{1}{2})^2$
$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Option (b) $\frac{1}{2}$
Q4
2025
00:00
If $x\left(\frac{2 \tan 30^{\circ}}{1 + \tan^2 30^{\circ}}\right) = y\left(\frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}}\right)$ then $x : y =$
LHS uses identity $\sin 2A$: $x(\sin 60^{\circ}) = x\frac{\sqrt{3}}{2}$
RHS uses identity $\tan 2A$: $y(\tan 60^{\circ}) = y\sqrt{3}$
$x\frac{\sqrt{3}}{2} = y\sqrt{3}$
$\frac{x}{2} = y \implies \frac{x}{y} = \frac{2}{1}$
Option (c) 2:1
Q5
2024C
00:00
If $\tan^2 \theta + \cot^2 \alpha = 2$, where $\theta = 45^{\circ}$ and $0^{\circ} \le \alpha \le 90^{\circ}$, then the value of $\alpha$ is :
Substitute $\theta = 45^{\circ}$: $\tan 45^{\circ} = 1$
$(1)^2 + \cot^2 \alpha = 2 \implies \cot^2 \alpha = 1$
$\cot \alpha = 1$ (since $\alpha$ is acute)
$\alpha = 45^{\circ}$
Option (b) $45^{\circ}$
Q6
2024
00:00
If $\cos (\alpha + \beta) = 0$, then value of $\cos ((\alpha + \beta)/2)$ is equal to :
$\cos(\alpha+\beta) = 0 \implies \alpha+\beta = 90^{\circ}$
Required: $\cos(\frac{90^{\circ}}{2}) = \cos 45^{\circ}$
Value is $\frac{1}{\sqrt{2}}$
Option (a) $\frac{1}{\sqrt{2}}$
Q7
2024
00:00
If $\cos \theta = \frac{\sqrt{3}}{2}$ and $\sin \phi = \frac{1}{2}$, then $\tan (\theta + \phi)$ is :
$\cos \theta = \frac{\sqrt{3}}{2} \implies \theta = 30^{\circ}$
$\sin \phi = \frac{1}{2} \implies \phi = 30^{\circ}$
$\theta + \phi = 60^{\circ}$
$\tan 60^{\circ} = \sqrt{3}$
Option (a) $\sqrt{3}$
Q8
2023
00:00
$\frac{2 \tan 30^{\circ}}{1 + \tan^2 30^{\circ}}$ is equal to :
Identity: $\frac{2\tan A}{1+\tan^2 A} = \sin 2A$
Here $A = 30^{\circ}$, so $\sin(2 \times 30^{\circ}) = \sin 60^{\circ}$
Alternatively calculate: $\frac{2(1/\sqrt{3})}{1+1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{\sqrt{3}}{2}$
Option (a) $\sin 60^{\circ}$
Q9
2023
00:00
$[ \frac{5}{8} \sec^2 60^{\circ} - \tan^2 60^{\circ} + \cos^2 45^{\circ} ]$ is equal to
Values: $\sec 60^{\circ}=2, \tan 60^{\circ}=\sqrt{3}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$
Substitute: $\frac{5}{8}(4) - 3 + \frac{1}{2}$
$\frac{5}{2} - 3 + \frac{1}{2} = \frac{6}{2} - 3 = 3 - 3 = 0$
Option (c) 0
Q10
2021-22
00:00
Given that $\sin \alpha = \frac{\sqrt{3}}{2}$ and $\tan \beta = \frac{1}{\sqrt{3}}$, then the value of $\cos (\alpha - \beta)$ is
$\sin \alpha = \frac{\sqrt{3}}{2} \implies \alpha = 60^{\circ}$
$\tan \beta = \frac{1}{\sqrt{3}} \implies \beta = 30^{\circ}$
$\alpha - \beta = 30^{\circ}$
$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$
Option (a) $\frac{\sqrt{3}}{2}$
Q11
2021-22
00:00
The value of $\theta$ for which $2 \sin 2\theta = 1$, is
$\sin 2\theta = \frac{1}{2}$
$2\theta = 30^{\circ}$
$\theta = 15^{\circ}$
Option (a) $15^{\circ}$
Q12
2020
00:00
Evaluate : $2 \sec 30^{\circ} \times \tan 60^{\circ}$
Values: $\sec 30^{\circ} = \frac{2}{\sqrt{3}}, \tan 60^{\circ} = \sqrt{3}$
$2 \times \frac{2}{\sqrt{3}} \times \sqrt{3} = 4$
4
Q13
2020
00:00
Write the value of $\sin^2 30^{\circ} + \cos^2 60^{\circ}$.
$\sin 30^{\circ} = \frac{1}{2}, \cos 60^{\circ} = \frac{1}{2}$
$(\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
$\frac{1}{2}$
Q14
2020
00:00
Evaluate : $\frac{2 \tan 45^{\circ} \times \cos 60^{\circ}}{\sin 30^{\circ}}$
Values: $\tan 45^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \sin 30^{\circ}=\frac{1}{2}$
Numerator: $2(1)(\frac{1}{2}) = 1$
Denominator: $\frac{1}{2}$
Result: $1 / (1/2) = 2$
2
Q15
2019
00:00
If $\sin x + \cos y = 1; x = 30^{\circ}$ and $y$ is an acute angle, find the value of $y$.
Substitute $x=30^{\circ}: \frac{1}{2} + \cos y = 1$
$\cos y = 1 - \frac{1}{2} = \frac{1}{2}$
$\cos y = \cos 60^{\circ} \implies y = 60^{\circ}$
$60^{\circ}$
Q16
2025
00:00
If $4k = \tan^2 60^{\circ} - 2\csc^2 30^{\circ} - 2\tan^2 30^{\circ}$, then find the value of k.
Values: $\tan 60^{\circ}=\sqrt{3}, \csc 30^{\circ}=2, \tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$4k = 3 - 2(4) - 2(\frac{1}{3})$
$4k = 3 - 8 - \frac{2}{3} = -5 - \frac{2}{3} = -\frac{17}{3}$
$k = -\frac{17}{12}$
$-\frac{17}{12}$
Q17
2025
00:00
If $x \cos 60^{\circ} + y \cos 0^{\circ} + \sin 30^{\circ} - \cot 45^{\circ} = 5$, then find the value of $x + 2y$.
Substitute: $x(\frac{1}{2}) + y(1) + \frac{1}{2} - 1 = 5$
$\frac{x}{2} + y - \frac{1}{2} = 5$
Multiply by 2: $x + 2y - 1 = 10$
$x + 2y = 11$
11
Q18
2025
00:00
Evaluate : $\frac{\tan^2 60^{\circ}}{\sin^2 60^{\circ} + \cos^2 30^{\circ}}$
Numerator: $(\sqrt{3})^2 = 3$
Denominator: $(\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} + \frac{3}{4} = \frac{3}{2}$
Result: $3 \div \frac{3}{2} = 2$
2
Q19
2024C
00:00
If $\cos (A + B) = \frac{1}{2}$ and $\tan (A - B) = \frac{1}{\sqrt{3}}$, where $0 \le A + B \le 90^{\circ}$, then find the value of $\sec (2A - 3B)$.
$\cos(A+B)=1/2 \implies A+B=60^{\circ}$
$\tan(A-B)=1/\sqrt{3} \implies A-B=30^{\circ}$
Solve: $2A=90^{\circ} \implies A=45^{\circ}, B=15^{\circ}$
$2A - 3B = 90^{\circ} - 45^{\circ} = 45^{\circ}$
$\sec 45^{\circ} = \sqrt{2}$
$\sqrt{2}$
Q20
2024C
00:00
Find the value of x such that, $3 \tan^2 60^{\circ} - x \sin^2 45^{\circ} + \frac{3}{4} \sec^2 30^{\circ} = 2 \csc^2 30^{\circ}$
$3(3) - x(\frac{1}{2}) + \frac{3}{4}(\frac{4}{3}) = 2(4)$
$9 - \frac{x}{2} + 1 = 8$
$10 - \frac{x}{2} = 8 \implies \frac{x}{2} = 2$
$x = 4$
4
Q21
2024
00:00
If $2 \sin(A + B) = \sqrt{3}$ and $\cos (A - B) = 1$, then find the measures of angles A and B. $0 \le A, B, (A + B) \le 90^{\circ}$.
$\sin(A+B) = \frac{\sqrt{3}}{2} \implies A+B=60^{\circ}$
$\cos(A-B) = 1 \implies A-B=0^{\circ}$
Solving gives $A = 30^{\circ}$ and $B = 30^{\circ}$
$A=30^{\circ}, B=30^{\circ}$
Q22
2024
00:00
Evaluate : $2\sqrt{2} \cos 45^{\circ} \sin 30^{\circ} + 2\sqrt{3} \cos 30^{\circ}$
$2\sqrt{2}(\frac{1}{\sqrt{2}})(\frac{1}{2}) + 2\sqrt{3}(\frac{\sqrt{3}}{2})$
$= 1 + 3 = 4$
4
Q23
2024
00:00
If $A = 60^{\circ}$ and $B = 30^{\circ}$, verify that : $\sin (A + B) = \sin A \cos B + \cos A \sin B$
LHS: $\sin(60+30) = \sin 90^{\circ} = 1$
RHS: $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2}$
$= \frac{3}{4} + \frac{1}{4} = 1$
LHS = RHS
Verified
Q24
2024
00:00
Evaluate : $\dfrac{\cos 45^{\circ} + \sin 60^{\circ}}{\sec 30^{\circ} + \csc 30^{\circ}}$
Numerator: $\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} = \frac{\sqrt{2}+\sqrt{3}}{2}$
Denominator: $\frac{2}{\sqrt{3}} + 2 = \frac{2+2\sqrt{3}}{\sqrt{3}}$
Result: $\frac{\sqrt{3}(\sqrt{2}+\sqrt{3})}{4(1+\sqrt{3})}$
$\frac{\sqrt{6}+3}{4(1+\sqrt{3})}$
Q25
2024
00:00
Evaluate : $\dfrac{2 \tan 30^{\circ} \cdot \sec 60^{\circ} \cdot \tan 45^{\circ}}{1 - \sin^2 60^{\circ}}$
Numerator: $2(\frac{1}{\sqrt{3}})(2)(1) = \frac{4}{\sqrt{3}}$
Denominator: $1 - \frac{3}{4} = \frac{1}{4}$
Result: $\frac{4}{\sqrt{3}} \times 4 = \frac{16\sqrt{3}}{3}$
$\frac{16\sqrt{3}}{3}$
Q26
2023
00:00
If $4 \cot^2 45^{\circ} - \sec^2 60^{\circ} + \sin^2 60^{\circ} + p = \frac{3}{4}$, then find the value of p.
$4(1)^2 - (2)^2 + (\frac{\sqrt{3}}{2})^2 + p = \frac{3}{4}$
$4 - 4 + \frac{3}{4} + p = \frac{3}{4}$
$p = 0$
0
Q27
2023
00:00
Evaluate $2 \sec^2 \theta + 3 \csc^2 \theta - 2 \sin \theta \cos \theta$ if $\theta = 45^{\circ}$.
Values: $\sec 45^{\circ}=\sqrt{2}, \csc 45^{\circ}=\sqrt{2}, \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$2(2) + 3(2) - 2(\frac{1}{2})$
$4 + 6 - 1 = 9$
9
Q28
2023
00:00
If $\sin \theta - \cos \theta = 0$, then find the value of $\sin^4 \theta + \cos^4 \theta$.
$\sin \theta = \cos \theta \implies \theta = 45^{\circ}$
$(\frac{1}{\sqrt{2}})^4 + (\frac{1}{\sqrt{2}})^4$
$\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
$\frac{1}{2}$
Q29
2023
00:00
Evaluate : $5/\cot^2 30^{\circ} + 1/\sin^2 60^{\circ} - \cot^2 45^{\circ} + 2 \sin^2 90^{\circ}$
$\frac{5}{3} + \frac{1}{3/4} - 1 + 2(1)$
$\frac{5}{3} + \frac{4}{3} + 1$
$\frac{9}{3} + 1 = 4$
4
Q30
2023
00:00
If $\theta$ is an acute angle and $\sin \theta = \cos \theta$, find the value of $\tan^2 \theta + \cot^2 \theta - 2$.
$\theta = 45^{\circ}$
$1^2 + 1^2 - 2 = 0$
0
Q31
2017
00:00
Take $A = 60^{\circ}$ and $B = 30^{\circ}$. Write the values of $\cos A + \cos B$ and $\cos(A + B)$. Is $\cos(A + B) = \cos A + \cos B$?
$\cos A + \cos B = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}$
$\cos(A+B) = \cos 90^{\circ} = 0$
Values are not equal.
No
Q32
2017
00:00
Find $\csc 30^{\circ}$ and $\cos 60^{\circ}$ geometrically.
Draw equilateral triangle ABC of side 2a
Draw altitude AD, splitting base into a, a
In $\Delta ABD$, $\angle B = 60^{\circ}, \angle BAD = 30^{\circ}$
$\cos 60^{\circ} = BD/AB = a/2a = 1/2$
$\sin 30^{\circ} = BD/AB = 1/2 \implies \csc 30^{\circ} = 2$
$\csc 30^{\circ}=2, \cos 60^{\circ}=1/2$
Q33
2017
00:00
$\sin(A + B) = 1$ & $\sin (A - B) = \frac{1}{2}$, $0 \le A + B = 90^{\circ}$ & $A > B$, then find A & B.
$A+B=90^{\circ}$
$A-B=30^{\circ}$
Adding equations: $2A = 120^{\circ} \implies A = 60^{\circ}$
$B = 30^{\circ}$
$A=60^{\circ}, B=30^{\circ}$
Q34
2017
00:00
If $\theta = 30^{\circ}$, verify the following : (i) $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$ (ii) $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$
(i) LHS $\cos 90=0$; RHS $4(\frac{\sqrt{3}}{2})^3 - 3(\frac{\sqrt{3}}{2}) = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0$
(ii) LHS $\sin 90=1$; RHS $3(\frac{1}{2}) - 4(\frac{1}{8}) = \frac{3}{2} - \frac{1}{2} = 1$
Verified
Q35
2017
00:00
Find trigonometric ratios of $30^{\circ}$ & $45^{\circ}$ in all values of T.R.
For $30^{\circ}$: $\sin=1/2, \cos=\sqrt{3}/2, \tan=1/\sqrt{3}$
For $45^{\circ}$: $\sin=1/\sqrt{2}, \cos=1/\sqrt{2}, \tan=1$
Reciprocals follow accordingly
Values listed
Q36
2016
00:00
If $\sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B$ and $\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B$. Find the value of (i) $\sin 75^{\circ}$ (ii) $\cos 15^{\circ}$
(i) $\sin(45+30) = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$
(ii) $\cos(45-30) = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$
$\frac{\sqrt{3}+1}{2\sqrt{2}}$ for both
If $\tan 3\theta = \sqrt{3}$, then $\theta/2$ equals
Given: $\tan 3\theta = \sqrt{3}$
We know that $\tan 60^{\circ} = \sqrt{3}$
So, $3\theta = 60^{\circ} \implies \theta = 20^{\circ}$
Required value: $\theta/2 = 20^{\circ}/2 = 10^{\circ}$
Option (d) $10^{\circ}$
Q2
2025
00:00
If $\sin 4\theta = \frac{\sqrt{3}}{2}$, then $\theta/3$ equals :
Given: $\sin 4\theta = \frac{\sqrt{3}}{2}$
We know that $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$
So, $4\theta = 60^{\circ} \implies \theta = 15^{\circ}$
Required value: $\theta/3 = 15^{\circ}/3 = 5^{\circ}$
Option (d) $5^{\circ}$
Q3
2025
00:00
If $\alpha + \beta = 90^{\circ}$ and $\alpha = 2\beta$, then $\cos^2 \alpha + \sin^2 \beta$ is equal to :
Substitute $\alpha = 2\beta$ into $\alpha + \beta = 90^{\circ}$
$3\beta = 90^{\circ} \implies \beta = 30^{\circ}$
Then $\alpha = 60^{\circ}$
Expression: $\cos^2 60^{\circ} + \sin^2 30^{\circ} = (\frac{1}{2})^2 + (\frac{1}{2})^2$
$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Option (b) $\frac{1}{2}$
Q4
2025
00:00
If $x\left(\frac{2 \tan 30^{\circ}}{1 + \tan^2 30^{\circ}}\right) = y\left(\frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}}\right)$ then $x : y =$
LHS uses identity $\sin 2A$: $x(\sin 60^{\circ}) = x\frac{\sqrt{3}}{2}$
RHS uses identity $\tan 2A$: $y(\tan 60^{\circ}) = y\sqrt{3}$
$x\frac{\sqrt{3}}{2} = y\sqrt{3}$
$\frac{x}{2} = y \implies \frac{x}{y} = \frac{2}{1}$
Option (c) 2:1
Q5
2024C
00:00
If $\tan^2 \theta + \cot^2 \alpha = 2$, where $\theta = 45^{\circ}$ and $0^{\circ} \le \alpha \le 90^{\circ}$, then the value of $\alpha$ is :
Substitute $\theta = 45^{\circ}$: $\tan 45^{\circ} = 1$
$(1)^2 + \cot^2 \alpha = 2 \implies \cot^2 \alpha = 1$
$\cot \alpha = 1$ (since $\alpha$ is acute)
$\alpha = 45^{\circ}$
Option (b) $45^{\circ}$
Q6
2024
00:00
If $\cos (\alpha + \beta) = 0$, then value of $\cos ((\alpha + \beta)/2)$ is equal to :
$\cos(\alpha+\beta) = 0 \implies \alpha+\beta = 90^{\circ}$
Required: $\cos(\frac{90^{\circ}}{2}) = \cos 45^{\circ}$
Value is $\frac{1}{\sqrt{2}}$
Option (a) $\frac{1}{\sqrt{2}}$
Q7
2024
00:00
If $\cos \theta = \frac{\sqrt{3}}{2}$ and $\sin \phi = \frac{1}{2}$, then $\tan (\theta + \phi)$ is :
$\cos \theta = \frac{\sqrt{3}}{2} \implies \theta = 30^{\circ}$
$\sin \phi = \frac{1}{2} \implies \phi = 30^{\circ}$
$\theta + \phi = 60^{\circ}$
$\tan 60^{\circ} = \sqrt{3}$
Option (a) $\sqrt{3}$
Q8
2023
00:00
$\frac{2 \tan 30^{\circ}}{1 + \tan^2 30^{\circ}}$ is equal to :
Identity: $\frac{2\tan A}{1+\tan^2 A} = \sin 2A$
Here $A = 30^{\circ}$, so $\sin(2 \times 30^{\circ}) = \sin 60^{\circ}$
Alternatively calculate: $\frac{2(1/\sqrt{3})}{1+1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{\sqrt{3}}{2}$
Option (a) $\sin 60^{\circ}$
Q9
2023
00:00
$[ \frac{5}{8} \sec^2 60^{\circ} - \tan^2 60^{\circ} + \cos^2 45^{\circ} ]$ is equal to
Values: $\sec 60^{\circ}=2, \tan 60^{\circ}=\sqrt{3}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$
Substitute: $\frac{5}{8}(4) - 3 + \frac{1}{2}$
$\frac{5}{2} - 3 + \frac{1}{2} = \frac{6}{2} - 3 = 3 - 3 = 0$
Option (c) 0
Q10
2021-22
00:00
Given that $\sin \alpha = \frac{\sqrt{3}}{2}$ and $\tan \beta = \frac{1}{\sqrt{3}}$, then the value of $\cos (\alpha - \beta)$ is
$\sin \alpha = \frac{\sqrt{3}}{2} \implies \alpha = 60^{\circ}$
$\tan \beta = \frac{1}{\sqrt{3}} \implies \beta = 30^{\circ}$
$\alpha - \beta = 30^{\circ}$
$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$
Option (a) $\frac{\sqrt{3}}{2}$
Q11
2021-22
00:00
The value of $\theta$ for which $2 \sin 2\theta = 1$, is
$\sin 2\theta = \frac{1}{2}$
$2\theta = 30^{\circ}$
$\theta = 15^{\circ}$
Option (a) $15^{\circ}$
Q12
2020
00:00
Evaluate : $2 \sec 30^{\circ} \times \tan 60^{\circ}$
Values: $\sec 30^{\circ} = \frac{2}{\sqrt{3}}, \tan 60^{\circ} = \sqrt{3}$
$2 \times \frac{2}{\sqrt{3}} \times \sqrt{3} = 4$
4
Q13
2020
00:00
Write the value of $\sin^2 30^{\circ} + \cos^2 60^{\circ}$.
$\sin 30^{\circ} = \frac{1}{2}, \cos 60^{\circ} = \frac{1}{2}$
$(\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
$\frac{1}{2}$
Q14
2020
00:00
Evaluate : $\frac{2 \tan 45^{\circ} \times \cos 60^{\circ}}{\sin 30^{\circ}}$
Values: $\tan 45^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \sin 30^{\circ}=\frac{1}{2}$
Numerator: $2(1)(\frac{1}{2}) = 1$
Denominator: $\frac{1}{2}$
Result: $1 / (1/2) = 2$
2
Q15
2019
00:00
If $\sin x + \cos y = 1; x = 30^{\circ}$ and $y$ is an acute angle, find the value of $y$.
Substitute $x=30^{\circ}: \frac{1}{2} + \cos y = 1$
$\cos y = 1 - \frac{1}{2} = \frac{1}{2}$
$\cos y = \cos 60^{\circ} \implies y = 60^{\circ}$
$60^{\circ}$
Q16
2025
00:00
If $4k = \tan^2 60^{\circ} - 2\csc^2 30^{\circ} - 2\tan^2 30^{\circ}$, then find the value of k.
Values: $\tan 60^{\circ}=\sqrt{3}, \csc 30^{\circ}=2, \tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$4k = 3 - 2(4) - 2(\frac{1}{3})$
$4k = 3 - 8 - \frac{2}{3} = -5 - \frac{2}{3} = -\frac{17}{3}$
$k = -\frac{17}{12}$
$-\frac{17}{12}$
Q17
2025
00:00
If $x \cos 60^{\circ} + y \cos 0^{\circ} + \sin 30^{\circ} - \cot 45^{\circ} = 5$, then find the value of $x + 2y$.
Substitute: $x(\frac{1}{2}) + y(1) + \frac{1}{2} - 1 = 5$
$\frac{x}{2} + y - \frac{1}{2} = 5$
Multiply by 2: $x + 2y - 1 = 10$
$x + 2y = 11$
11
Q18
2025
00:00
Evaluate : $\frac{\tan^2 60^{\circ}}{\sin^2 60^{\circ} + \cos^2 30^{\circ}}$
Numerator: $(\sqrt{3})^2 = 3$
Denominator: $(\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} + \frac{3}{4} = \frac{3}{2}$
Result: $3 \div \frac{3}{2} = 2$
2
Q19
2024C
00:00
If $\cos (A + B) = \frac{1}{2}$ and $\tan (A - B) = \frac{1}{\sqrt{3}}$, where $0 \le A + B \le 90^{\circ}$, then find the value of $\sec (2A - 3B)$.
$\cos(A+B)=1/2 \implies A+B=60^{\circ}$
$\tan(A-B)=1/\sqrt{3} \implies A-B=30^{\circ}$
Solve: $2A=90^{\circ} \implies A=45^{\circ}, B=15^{\circ}$
$2A - 3B = 90^{\circ} - 45^{\circ} = 45^{\circ}$
$\sec 45^{\circ} = \sqrt{2}$
$\sqrt{2}$
Q20
2024C
00:00
Find the value of x such that, $3 \tan^2 60^{\circ} - x \sin^2 45^{\circ} + \frac{3}{4} \sec^2 30^{\circ} = 2 \csc^2 30^{\circ}$
$3(3) - x(\frac{1}{2}) + \frac{3}{4}(\frac{4}{3}) = 2(4)$
$9 - \frac{x}{2} + 1 = 8$
$10 - \frac{x}{2} = 8 \implies \frac{x}{2} = 2$
$x = 4$
4
Q21
2024
00:00
If $2 \sin(A + B) = \sqrt{3}$ and $\cos (A - B) = 1$, then find the measures of angles A and B. $0 \le A, B, (A + B) \le 90^{\circ}$.
$\sin(A+B) = \frac{\sqrt{3}}{2} \implies A+B=60^{\circ}$
$\cos(A-B) = 1 \implies A-B=0^{\circ}$
Solving gives $A = 30^{\circ}$ and $B = 30^{\circ}$
$A=30^{\circ}, B=30^{\circ}$
Q22
2024
00:00
Evaluate : $2\sqrt{2} \cos 45^{\circ} \sin 30^{\circ} + 2\sqrt{3} \cos 30^{\circ}$
$2\sqrt{2}(\frac{1}{\sqrt{2}})(\frac{1}{2}) + 2\sqrt{3}(\frac{\sqrt{3}}{2})$
$= 1 + 3 = 4$
4
Q23
2024
00:00
If $A = 60^{\circ}$ and $B = 30^{\circ}$, verify that : $\sin (A + B) = \sin A \cos B + \cos A \sin B$
LHS: $\sin(60+30) = \sin 90^{\circ} = 1$
RHS: $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2}$
$= \frac{3}{4} + \frac{1}{4} = 1$
LHS = RHS
Verified
Q24
2024
00:00
Evaluate : $\dfrac{\cos 45^{\circ} + \sin 60^{\circ}}{\sec 30^{\circ} + \csc 30^{\circ}}$
Numerator: $\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} = \frac{\sqrt{2}+\sqrt{3}}{2}$
Denominator: $\frac{2}{\sqrt{3}} + 2 = \frac{2+2\sqrt{3}}{\sqrt{3}}$
Result: $\frac{\sqrt{3}(\sqrt{2}+\sqrt{3})}{4(1+\sqrt{3})}$
$\frac{\sqrt{6}+3}{4(1+\sqrt{3})}$
Q25
2024
00:00
Evaluate : $\dfrac{2 \tan 30^{\circ} \cdot \sec 60^{\circ} \cdot \tan 45^{\circ}}{1 - \sin^2 60^{\circ}}$
Numerator: $2(\frac{1}{\sqrt{3}})(2)(1) = \frac{4}{\sqrt{3}}$
Denominator: $1 - \frac{3}{4} = \frac{1}{4}$
Result: $\frac{4}{\sqrt{3}} \times 4 = \frac{16\sqrt{3}}{3}$
$\frac{16\sqrt{3}}{3}$
Q26
2023
00:00
If $4 \cot^2 45^{\circ} - \sec^2 60^{\circ} + \sin^2 60^{\circ} + p = \frac{3}{4}$, then find the value of p.
$4(1)^2 - (2)^2 + (\frac{\sqrt{3}}{2})^2 + p = \frac{3}{4}$
$4 - 4 + \frac{3}{4} + p = \frac{3}{4}$
$p = 0$
0
Q27
2023
00:00
Evaluate $2 \sec^2 \theta + 3 \csc^2 \theta - 2 \sin \theta \cos \theta$ if $\theta = 45^{\circ}$.
Values: $\sec 45^{\circ}=\sqrt{2}, \csc 45^{\circ}=\sqrt{2}, \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$2(2) + 3(2) - 2(\frac{1}{2})$
$4 + 6 - 1 = 9$
9
Q28
2023
00:00
If $\sin \theta - \cos \theta = 0$, then find the value of $\sin^4 \theta + \cos^4 \theta$.
$\sin \theta = \cos \theta \implies \theta = 45^{\circ}$
$(\frac{1}{\sqrt{2}})^4 + (\frac{1}{\sqrt{2}})^4$
$\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
$\frac{1}{2}$
Q29
2023
00:00
Evaluate : $5/\cot^2 30^{\circ} + 1/\sin^2 60^{\circ} - \cot^2 45^{\circ} + 2 \sin^2 90^{\circ}$
$\frac{5}{3} + \frac{1}{3/4} - 1 + 2(1)$
$\frac{5}{3} + \frac{4}{3} + 1$
$\frac{9}{3} + 1 = 4$
4
Q30
2023
00:00
If $\theta$ is an acute angle and $\sin \theta = \cos \theta$, find the value of $\tan^2 \theta + \cot^2 \theta - 2$.
$\theta = 45^{\circ}$
$1^2 + 1^2 - 2 = 0$
0
Q31
2017
00:00
Take $A = 60^{\circ}$ and $B = 30^{\circ}$. Write the values of $\cos A + \cos B$ and $\cos(A + B)$. Is $\cos(A + B) = \cos A + \cos B$?
$\cos A + \cos B = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}$
$\cos(A+B) = \cos 90^{\circ} = 0$
Values are not equal.
No
Q32
2017
00:00
Find $\csc 30^{\circ}$ and $\cos 60^{\circ}$ geometrically.
Draw equilateral triangle ABC of side 2a
Draw altitude AD, splitting base into a, a
In $\Delta ABD$, $\angle B = 60^{\circ}, \angle BAD = 30^{\circ}$
$\cos 60^{\circ} = BD/AB = a/2a = 1/2$
$\sin 30^{\circ} = BD/AB = 1/2 \implies \csc 30^{\circ} = 2$
$\csc 30^{\circ}=2, \cos 60^{\circ}=1/2$
Q33
2017
00:00
$\sin(A + B) = 1$ & $\sin (A - B) = \frac{1}{2}$, $0 \le A + B = 90^{\circ}$ & $A > B$, then find A & B.
$A+B=90^{\circ}$
$A-B=30^{\circ}$
Adding equations: $2A = 120^{\circ} \implies A = 60^{\circ}$
$B = 30^{\circ}$
$A=60^{\circ}, B=30^{\circ}$
Q34
2017
00:00
If $\theta = 30^{\circ}$, verify the following : (i) $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$ (ii) $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$
(i) LHS $\cos 90=0$; RHS $4(\frac{\sqrt{3}}{2})^3 - 3(\frac{\sqrt{3}}{2}) = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0$
(ii) LHS $\sin 90=1$; RHS $3(\frac{1}{2}) - 4(\frac{1}{8}) = \frac{3}{2} - \frac{1}{2} = 1$
Verified
Q35
2017
00:00
Find trigonometric ratios of $30^{\circ}$ & $45^{\circ}$ in all values of T.R.
For $30^{\circ}$: $\sin=1/2, \cos=\sqrt{3}/2, \tan=1/\sqrt{3}$
For $45^{\circ}$: $\sin=1/\sqrt{2}, \cos=1/\sqrt{2}, \tan=1$
Reciprocals follow accordingly
Values listed
Q36
2016
00:00
If $\sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B$ and $\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B$. Find the value of (i) $\sin 75^{\circ}$ (ii) $\cos 15^{\circ}$
(i) $\sin(45+30) = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$
(ii) $\cos(45-30) = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$
$\frac{\sqrt{3}+1}{2\sqrt{2}}$ for both