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.

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.$4+\sqrt{10}$
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   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
   If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$, then $\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)$ is equal to
   1. A.$x-\tan ^{-1} \frac{4}{3}$
   2. B.$x+\tan ^{-1} \frac{4}{5}$
   3. C.$x-\tan ^{-1} \frac{5}{12}$
   4. D.$x+\tan ^{-1} \frac{5}{12}$
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   Answer: (C)

   $\begin{aligned} & \frac{12}{13} \cos x+\frac{5}{13} \sin x \\ & \text { Let } \tan \alpha=\frac{5}{12}, \alpha \in\left(0, \frac{\pi}{2}\right) \\ & \Rightarrow \sin \alpha=\frac{5}{13}, \cos \alpha=\frac{12}{13}\end{aligned}$ $\begin{aligned} & \Rightarrow \frac{12}{13} \cos x+\frac{5}{13} \sin x=\cos \alpha \cos x+\sin \alpha \sin x \\ & \quad=\cos (x-\alpha) \\ & \Rightarrow \cos ^{-1}[\cos (x-\alpha)]=x-\alpha \\ & \quad=x-\tan ^{-1}\left(\frac{5}{12}\right)\end{aligned}$
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
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   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.$2$
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   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.$\sum _{\theta \in S}\theta =\dfrac{3\pi }{4}$
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   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, $\sum _{\theta \in S}\theta =\dfrac{\pi }{6}+\dfrac{\pi }{3}=\dfrac{\pi }{2}$