Trigonometric Ratios & Identities JEE Main previous year questions with solutions

5 solved JEE Main questions on Trigonometric Ratios & Identities, free to read — no sign-in needed. The full chapter has 66 questions; sign in to attempt the remaining 61 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (23 Jan, Shift 2)Maximum and Minimum Values
   The least value of $\left(\cos ^{2} \theta-6 \sin \theta \cos \theta+3 \sin ^{2} \theta+2\right)$ is
   1. A.$-1$
   2. B.1
   3. C.$4-\sqrt{10}$
   4. D.$4+\sqrt{10}$
   Show answer & solution

   Answer: (C)

   $\cos^2\theta + 3\sin^2\theta - 3\sin 2\theta + 2$. Using double angle formulas: $= \frac{1+\cos 2\theta}{2} + \frac{3(1-\cos 2\theta)}{2} - 3\sin 2\theta + 2$ $= 4 - \cos 2\theta - 3\sin 2\theta$. Minimum of $-\cos 2\theta - 3\sin 2\theta = -\sqrt{1+9} = -\sqrt{10}$. Least value $= 4 - \sqrt{10}$.
2. Q2JEE Main 2025 (23 Jan, Shift 1)Transformation Formulas
   The value of $\left(\sin 70^{\circ}\right)\left(\cot 10^{\circ} \cot 70^{\circ}-1\right)$ is
   1. A.$2 / 3$
   2. B.$1$
   3. C.$0$
   4. D.$3 / 2$
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   Answer: (B)

   $\begin{aligned} & \left(\sin 70^{\circ}\right)\left(\cot 10^{\circ} \cot 70^{\circ}-1\right) \\ & =\sin 70^{\circ} \cot 10^{\circ} \cot 70^{\circ}-\sin 70^{\circ} \\ & =\cot 10^{\circ} \cos 70^{\circ}-\sin 70^{\circ} \\ & =\frac{\cos 10^{\circ} \cos 70^{\circ}-\sin 70^{\circ} \sin 10^{\circ}}{\sin 10^{\circ}} \\ & =\frac{\cos \left(10^{\circ}+70^{\circ}\right)}{\sin 10^{\circ}} \\ & =\frac{\cos 80^{\circ}}{\sin 10^{\circ}}=1\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 1)Basic Identities & T Ratios
   If $\sin x=-\frac{3}{5}$, where $\pi \lt x \lt \frac{3 \pi}{2}$, then $80\left(\tan ^2 x-\cos x\right)$ is equal to
   1. A.108
   2. B.109
   3. C.18
   4. D.19
   Show answer & solution

   Answer: (B)

   $\begin{aligned} & \sin x=\frac{-3}{5}, \pi \lt x \lt \frac{3 \pi}{2} \\ & \tan x=\frac{3}{4} \cos x=-\frac{4}{5} \\ & 80\left(\tan ^2 x-\cos x\right) \\ & =80\left(\frac{9}{16}+\frac{4}{5}\right)=45+64=109\end{aligned}$
4. Q4JEE Main 2023 (10 Apr, Shift 1)Trigonometric Series
   $96\cos \dfrac{\pi }{33}\cos \dfrac{2\pi }{33}\cos \dfrac{4\pi }{33}\cos \dfrac{8\pi }{33}\cos \dfrac{16\pi }{33}$ is equal to
   1. A.$3$
   2. B.$1$
   3. C.$4$
   4. D.$2$
   Show answer & solution

   Answer: (A)

   Given, Expression $96\cdot \cos \dfrac{\pi }{33}\cdot \cos \dfrac{2\pi }{33}\cdot \cos \dfrac{4\pi }{33}.........\cos \dfrac{16\pi }{33}$ Now we know that, $\cos A\cdot \cos 2A\cdot \cos {2}^{2}A\cdot \cos {2}^{3}A.....\cdot \cos {2}^{n-1}A=\dfrac{\sin {2}^{n}A}{{2}^{n}\sin A}$ Now using the above formula in given expression we get, $96\cdot \cos \dfrac{\pi }{33}\cdot \cos \dfrac{2\pi }{33}\cdot \cos \dfrac{4\pi }{33}.........\cos \dfrac{16\pi }{33}$ $=96\times \dfrac{\sin \dfrac{32\pi }{33}}{{2}^{5}\sin \dfrac{\pi }{33}}$ $=96\times \dfrac{\sin \left(\pi -\dfrac{\pi }{33}\right)}{{2}^{5}\sin \dfrac{\pi }{33}}$ $=96\times \dfrac{\sin \left(\dfrac{\pi }{33}\right)}{{2}^{5}\sin \dfrac{\pi }{33}}\left\{\text{as}\sin \left(\pi -\alpha \right)=\sin \alpha \right\}$ $=96\times \dfrac{1}{32}=3$
5. Q5JEE Main 2022 (27 Jul, Shift 2)Sum and Difference & Multiple angle Formula
   Let $S=\left\{\theta \in \left(0,\dfrac{\pi }{2}\right):\sum _{m=1}^{9}\sec \left(\theta +\left(m-1\right)\dfrac{\pi }{6}\right)\sec \left(\theta +\dfrac{m\pi }{6}\right)=-\dfrac{8}{\sqrt{3}}\right\}$. Then
   1. A.$S=\left\{\dfrac{\pi }{12}\right\}$
   2. B.$S=\left\{\dfrac{2\pi }{3}\right\}$
   3. C.$\sum _{\theta \in S}\theta =\dfrac{\pi }{2}$
   4. D.$\sum _{\theta \in S}\theta =\dfrac{3\pi }{4}$
   Show answer & solution

   Answer: (C)

   Let $\theta +\left(m-1\right)\dfrac{\pi }{6}=x$ and $\theta +m\dfrac{\pi }{6}=y$ So, $y-x=\dfrac{\pi }{6}$ Now, $\sum _{m=1}^{9}\sec \left(\theta +\left(m-1\right)\dfrac{\pi }{6}\right)\sec \left(\theta +\dfrac{m\pi }{6}\right)$ $=\sum _{m=1}^{9}\sec x\sec y=\sum _{m=1}^{9}\dfrac{1}{\cos x\cos y}$ $=2\sum _{m=1}^{9}\dfrac{\sin \left(y-x\right)}{\cos x\cos y}=2\sum _{m=1}^{9}\left(\tan y-\tan x\right)$ $=2\sum _{m=1}^{9}\left(\tan \left(\theta +m\dfrac{\pi }{6}\right)-\tan \left(\theta +(m-1)\dfrac{\pi }{6}\right)\right)$ $=2\left(\tan \left(\theta +\dfrac{\pi }{6}\right)-\tan \left(\theta +\dfrac{0\cdot \pi }{6}\right)\right)+2\left(\tan \left(\theta +2\dfrac{\pi }{6}\right)-\tan \left(\theta +\dfrac{\pi }{6}\right)\right)$ $+2\left(\tan \left(\theta +3\dfrac{\pi }{6}\right)-\tan \left(\theta +2\dfrac{\pi }{6}\right)\right)+...+2\left(\tan \left(\theta +9\dfrac{\pi }{6}\right)-\tan \left(\theta +8\dfrac{\pi }{6}\right)\right)$ $=2\left(\tan \left(\theta +\dfrac{9\pi }{6}\right)-\tan \theta \right)=2\left(-\cot \theta -\tan \theta \right)$ i.e. $2\left(-\cot \theta -\tan \theta \right)=-\dfrac{8}{\sqrt{3}}$ (Given) $∴\tan \theta +\cot \theta =\dfrac{4}{\sqrt{3}}$ $\Rightarrow \tan \theta =\dfrac{1}{\sqrt{3}}$ or $\sqrt{3}$ (as $\theta \in \left(0,\dfrac{\pi }{2}\right)$) So, $S=\left\{\dfrac{\pi }{6},\dfrac{\pi }{3}\right\}$ Hence, $\sum _{\theta \in S}\theta =\dfrac{\pi }{6}+\dfrac{\pi }{3}=\dfrac{\pi }{2}$