Trigonometric Ratios PYQs
Overview
This page provides comprehensive Class 10 Maths Trigonometric Ratios PYQs | Introduction to Trigonometry. 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
2024
00:00
If $5\tan\theta-12=0$, then the value of $\sin\theta$ is:
Given: $5\tan\theta-12=0$
$5\tan\theta=12$
$\tan\theta=\frac{12}{5}$
Take opposite = 12, adjacent = 5
Hypotenuse $=\sqrt{12^2+5^2}=13$
$\sin\theta=\frac{12}{13}$
Option (b) $\frac{12}{13}$
Q2
2023
00:00
If $\tan\theta=\frac{5}{12}$, find $\dfrac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}$
Opposite = 5, Adjacent = 12
Hypotenuse = $13$
$\sin\theta=\frac{5}{13}, \cos\theta=\frac{12}{13}$
$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=\frac{17/13}{-7/13}=-\frac{17}{7}$
Option (a) $-\frac{17}{7}$
Q3
2023
00:00
If $2\tan A=3$, find $\dfrac{4\sin A+3\cos A}{4\sin A-3\cos A}$
$\tan A=\frac{3}{2}$
Opposite = 3, Adjacent = 2
Hypotenuse $=\sqrt{13}$
$\sin A=\frac{3}{\sqrt{13}}, \cos A=\frac{2}{\sqrt{13}}$
Substitute and simplify
Value $=3$
Option (c) $3$
Q4
2021-22
00:00
If $\cos\theta=\frac{\sqrt{3}}{2}$, find $\dfrac{\csc^2\theta-\sec^2\theta}{\csc^2\theta+\sec^2\theta}$
$\sin\theta=\frac{1}{2}$
$\sec^2\theta=\frac{4}{3}, \csc^2\theta=4$
Substitute and simplify
Value $=\frac{1}{2}$
Option (c) $\frac{1}{2}$
Q5
2021-22
00:00
Find the value of $\dfrac{1}{\csc\theta(1-\cot\theta)}+\dfrac{1}{\sec\theta(1-\tan\theta)}$
Convert into sine and cosine form
Simplify numerator and denominator
Expression becomes $\sin\theta+\cos\theta$
Option (c) $\sin\theta+\cos\theta$
Q6
2020
00:00
If $\sin\theta=\cos\theta$, find $\tan^2\theta+\cot^2\theta$
$\tan\theta=1, \cot\theta=1$
$\tan^2\theta+\cot^2\theta=2$
Option (a) $2$
Q7
2021
00:00
If $\tan\theta+\cot\theta=\frac{4\sqrt{3}}{3}$, find $\tan^2\theta+\cot^2\theta$
Square both sides
$\tan^2\theta+\cot^2\theta+2=\frac{16}{3}$
$\tan^2\theta+\cot^2\theta=\frac{10}{3}$
$\frac{10}{3}$
Q8
2017
00:00
If $\sin\alpha=\frac{1}{2}$, find $3\sin\alpha-4\sin^3\alpha$
$3\sin\alpha=\frac{3}{2}$
$4\sin^3\alpha=\frac{1}{2}$
Result $=1$
$1$
Q9
2025
00:00
If $\tan A+\cot A=6$, find $\tan^2 A+\cot^2 A-4$
Square: $\tan^2A+\cot^2A+2=36$
$\tan^2A+\cot^2A=34$
Required value $=30$
$30$
Q10
2020
00:00
If $15\cot A=8$, find $\sin A$ and $\sec A$
$\cot A=\frac{8}{15}$
Opposite = 15, Adjacent = 8, Hypotenuse = 17
$\sin A=\frac{15}{17}, \sec A=\frac{17}{8}$
$\sin A=\frac{15}{17},\; \sec A=\frac{17}{8}$
Q11
2024
00:00
Prove that $\dfrac{\tan\theta-\cot\theta}{\sin\theta\cos\theta}=\sec^2\theta-\csc^2\theta$
Convert into sine and cosine
Simplify numerator and denominator
Obtain $\sec^2\theta-\csc^2\theta$
Hence proved
Q12
2020
00:00
Prove that $\dfrac{1+\tan A}{2\sin A}+\dfrac{1+\cot A}{2\cos A}=\csc A+\sec A$
Convert into sine-cosine
Take LCM and simplify
RHS obtained
Hence proved
Q13
2019
00:00
Prove that $\dfrac{\tan\theta}{1-\tan\theta}-\dfrac{\cot\theta}{1-\cot\theta}=\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$
Simplify LHS using sine and cosine
Obtain RHS
Identity proved
Q14
2019
00:00
If $\cos\theta+\sin\theta=\sqrt{2}\cos\theta$, show that $\cos\theta-\sin\theta=\sqrt{2}\sin\theta$
Express $\sin\theta$ in terms of $\cos\theta$
Substitute and simplify
Hence proved
Q15
2016
00:00
If $\sin A=\frac{3}{5}$, find all other trigonometric ratios of $A$
Opposite = 3, Hypotenuse = 5
Adjacent = 4
Find remaining ratios
$\cos A=\frac{4}{5}, \tan A=\frac{3}{4}, \cot A=\frac{4}{3}, \sec A=\frac{5}{4}, \csc A=\frac{5}{3}$
Q16
2017
00:00
If $3\tan A=4$, verify $\dfrac{1-\tan^2 A}{1+\tan^2 A}=\cos^2 A-\sin^2 A$
$\tan A=\frac{4}{3}$
LHS $=-\frac{7}{25}$
$\sin A=\frac{4}{5}, \cos A=\frac{3}{5}$
RHS $=-\frac{7}{25}$
LHS = RHS, hence verified
If $5\tan\theta-12=0$, then the value of $\sin\theta$ is:
Given: $5\tan\theta-12=0$
$5\tan\theta=12$
$\tan\theta=\frac{12}{5}$
Take opposite = 12, adjacent = 5
Hypotenuse $=\sqrt{12^2+5^2}=13$
$\sin\theta=\frac{12}{13}$
Option (b) $\frac{12}{13}$
Q2
2023
00:00
If $\tan\theta=\frac{5}{12}$, find $\dfrac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}$
Opposite = 5, Adjacent = 12
Hypotenuse = $13$
$\sin\theta=\frac{5}{13}, \cos\theta=\frac{12}{13}$
$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=\frac{17/13}{-7/13}=-\frac{17}{7}$
Option (a) $-\frac{17}{7}$
Q3
2023
00:00
If $2\tan A=3$, find $\dfrac{4\sin A+3\cos A}{4\sin A-3\cos A}$
$\tan A=\frac{3}{2}$
Opposite = 3, Adjacent = 2
Hypotenuse $=\sqrt{13}$
$\sin A=\frac{3}{\sqrt{13}}, \cos A=\frac{2}{\sqrt{13}}$
Substitute and simplify
Value $=3$
Option (c) $3$
Q4
2021-22
00:00
If $\cos\theta=\frac{\sqrt{3}}{2}$, find $\dfrac{\csc^2\theta-\sec^2\theta}{\csc^2\theta+\sec^2\theta}$
$\sin\theta=\frac{1}{2}$
$\sec^2\theta=\frac{4}{3}, \csc^2\theta=4$
Substitute and simplify
Value $=\frac{1}{2}$
Option (c) $\frac{1}{2}$
Q5
2021-22
00:00
Find the value of $\dfrac{1}{\csc\theta(1-\cot\theta)}+\dfrac{1}{\sec\theta(1-\tan\theta)}$
Convert into sine and cosine form
Simplify numerator and denominator
Expression becomes $\sin\theta+\cos\theta$
Option (c) $\sin\theta+\cos\theta$
Q6
2020
00:00
If $\sin\theta=\cos\theta$, find $\tan^2\theta+\cot^2\theta$
$\tan\theta=1, \cot\theta=1$
$\tan^2\theta+\cot^2\theta=2$
Option (a) $2$
Q7
2021
00:00
If $\tan\theta+\cot\theta=\frac{4\sqrt{3}}{3}$, find $\tan^2\theta+\cot^2\theta$
Square both sides
$\tan^2\theta+\cot^2\theta+2=\frac{16}{3}$
$\tan^2\theta+\cot^2\theta=\frac{10}{3}$
$\frac{10}{3}$
Q8
2017
00:00
If $\sin\alpha=\frac{1}{2}$, find $3\sin\alpha-4\sin^3\alpha$
$3\sin\alpha=\frac{3}{2}$
$4\sin^3\alpha=\frac{1}{2}$
Result $=1$
$1$
Q9
2025
00:00
If $\tan A+\cot A=6$, find $\tan^2 A+\cot^2 A-4$
Square: $\tan^2A+\cot^2A+2=36$
$\tan^2A+\cot^2A=34$
Required value $=30$
$30$
Q10
2020
00:00
If $15\cot A=8$, find $\sin A$ and $\sec A$
$\cot A=\frac{8}{15}$
Opposite = 15, Adjacent = 8, Hypotenuse = 17
$\sin A=\frac{15}{17}, \sec A=\frac{17}{8}$
$\sin A=\frac{15}{17},\; \sec A=\frac{17}{8}$
Q11
2024
00:00
Prove that $\dfrac{\tan\theta-\cot\theta}{\sin\theta\cos\theta}=\sec^2\theta-\csc^2\theta$
Convert into sine and cosine
Simplify numerator and denominator
Obtain $\sec^2\theta-\csc^2\theta$
Hence proved
Q12
2020
00:00
Prove that $\dfrac{1+\tan A}{2\sin A}+\dfrac{1+\cot A}{2\cos A}=\csc A+\sec A$
Convert into sine-cosine
Take LCM and simplify
RHS obtained
Hence proved
Q13
2019
00:00
Prove that $\dfrac{\tan\theta}{1-\tan\theta}-\dfrac{\cot\theta}{1-\cot\theta}=\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$
Simplify LHS using sine and cosine
Obtain RHS
Identity proved
Q14
2019
00:00
If $\cos\theta+\sin\theta=\sqrt{2}\cos\theta$, show that $\cos\theta-\sin\theta=\sqrt{2}\sin\theta$
Express $\sin\theta$ in terms of $\cos\theta$
Substitute and simplify
Hence proved
Q15
2016
00:00
If $\sin A=\frac{3}{5}$, find all other trigonometric ratios of $A$
Opposite = 3, Hypotenuse = 5
Adjacent = 4
Find remaining ratios
$\cos A=\frac{4}{5}, \tan A=\frac{3}{4}, \cot A=\frac{4}{3}, \sec A=\frac{5}{4}, \csc A=\frac{5}{3}$
Q16
2017
00:00
If $3\tan A=4$, verify $\dfrac{1-\tan^2 A}{1+\tan^2 A}=\cos^2 A-\sin^2 A$
$\tan A=\frac{4}{3}$
LHS $=-\frac{7}{25}$
$\sin A=\frac{4}{5}, \cos A=\frac{3}{5}$
RHS $=-\frac{7}{25}$
LHS = RHS, hence verified