Trigonometric Functions JEE Main PYQs (2020–2026)

Solved Trigonometric Functions PYQs

3 recent questions with answers. The full set of 50 is free to practise.

1. JEE Main 2026HardCompound Angle Formulas (A +/- B)
   Q1

   Let $P=\{\theta \in [0,4\pi ]:{\tan }^2\theta \ne 1\}$ and $=\{a\in \mathbb{Z}:2({\cos }^8\theta -{\sin }^8\theta )sec2\theta =a^2,\theta \in P\}$ . Then $n(S)$ is:

   • A0
   • B1
   • C2
   • D3

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   Correct answer: (A)

   $$\begin{array}{rl}& 2\left({\cos }^8\theta -{\sin }^8\theta \right)\sec 2\theta =a^2\\ & 2\left({\cos }^4\theta +{\sin }^4\theta \right)\left({\cos }^2\theta +{\sin }^2\theta \right)\\ \left({\cos }^2\theta -{\sin }^2\theta \right)\sec 2\theta =a^2& \\ & 2\left({\cos }^4\theta +{\sin }^4\theta \right)=a^2\\ & 2\left(1-2{\sin }^2\theta {\cos }^2\theta \right)=a^2\\ & 2\left(1-\frac{{\sin }^{2}2\theta }{2}\right)=a^2,a\in Z\\ & a^2=2-{\sin }^{2}2\theta \in [1,2]\\ & a^2=1\text{ at }{\sin }^{2}2\theta =1\\ & 2\theta =(2n+1)\frac{\pi }{2}\\ & \theta =(2n+1)\frac{\pi }{4}\notin P\\ & n(S)=0\end{array}$$
2. JEE Main 2026HardMultiple and Submultiple Angle Formulas (2A, 3A, A/2)
   Q2

   The value of $\frac{\sqrt{3}cosec20^{\circ }-sec20^{\circ }}{\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{\circ }}$ is equal to

   • A32
   • B64
   • C12
   • D16

   Show answer & solution

   Correct answer: (B)

   $$\begin{array}{rl}& \begin{array}{rl}& \frac{\sqrt{3}cosec20^{\circ }-sec20^{\circ }}{\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{\circ }}\\ & =\frac{\frac{\sqrt{3}}{\sin 20^{\circ }}-\frac{1}{\cos 20^{\circ }}}{\cos 20^{\circ }\cos 40^{\circ }\cos 80^{\circ }\cdot \frac{1}{2}},\left\{\cos 60^{\circ }=\frac{1}{2}\right\}\\ & =\frac{2\left(\frac{\sqrt{3}\cos 20^{\circ }-sin20^{\circ }}{\sin 20^{\circ }\cos 20^{\circ }}\right)}{\cos 20^{\circ }\cos 40^{\circ }\cos 80^{\circ }},\\ =\frac{2\left(\sqrt{3}\cos 20^{\circ }-sin20^{\circ }\right)}{\frac{\sin 40^{\circ }}{2}\cdot cos20^{\circ }\cos 40^{\circ }\cos 80^{\circ }},& \\ \left\{\sin 20^{\circ }\cos 20^{\circ }=\frac{\sin 40^{\circ }}{2}\right\}& \\ & \text{ Numerator }\sqrt{3}\cos 20^{\circ }-sin20^{\circ }\\ =2\cdot \left(\frac{\sqrt{3}}{2}\cos 20^{\circ }-\frac{1}{2}\sin 20^{\circ }\right)& \\ =2\left(\sin 60^{\circ }\cos 20^{\circ }-cos60^{\circ }\sin 20^{\circ }\right)& \\ =2sin(60-20)=2\sin 40^{\circ }& \end{array}\\ & \text{ Substitute this value }\\ & \begin{array}{r}=\frac{2\cdot 2\sin 40^{\circ }}{\frac{\sin 80^{\circ }}{4}\cos 20^{\circ }\cos 80^{\circ }}=\frac{4\sin 40^{\circ }}{\frac{\sin 160^{\circ }}{8}\cos 20^{\circ }}\\ \left\{\sin 80^{\circ }\cos 80^{\circ }=\frac{\sin 160^{\circ }}{2}\right\}\\ =\frac{4\times 2\sin 20^{\circ }\cdot cos20^{\circ }}{\sin (180-160)\cos 20^{\circ }}\times 8\\ =\frac{64\sin 20^{\circ }}{\sin 20^{\circ }}=64\end{array}\end{array}$$
3. JEE Main 2026HardConditional Trigonometric Identities
   Q3

   Let $cos(\alpha +\beta )=-\frac{1}{10}$ and $sin(\alpha -\beta )=\frac{3}{8}$, where $0<\alpha <\frac{\pi }{3}$ and $0<\beta <\frac{\pi }{4}$. If $\tan 2\alpha =\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})},r,s\in N$, then $r+s$ is equal to $\_\_\_\_$.

   Numerical answer type

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   Correct answer: 20

   $$\begin{array}{rl}& \tan 2\alpha =tan[(\alpha +\beta )+(\alpha -\beta )]\\ & \tan 2\alpha =\frac{tan(\alpha +\beta )+tan(\alpha -\beta )}{1-tan(\alpha +\beta )\cdot tan(\alpha -\beta )}\\ & \tan 2\alpha =\frac{\left(-\sqrt{99}+\frac{3}{\sqrt{55}}\right)}{1-(\sqrt{99})\left(\frac{3}{\sqrt{55}}\right)}\\ & \tan 2\alpha =\frac{-3\sqrt{11}+\frac{3}{\sqrt{5}\times \sqrt{11}}}{1+\frac{9\sqrt{11}}{\sqrt{5}\times \sqrt{11}}}\\ & \tan 2\alpha =\frac{3(1-11\sqrt{5})}{\sqrt{11}(9+\sqrt{5})}\\ & r=11,s=9\\ & r+s=20\end{array}$$

Topic-wise question count

Where JEE Main has asked from Trigonometric Functions, most frequent first.

• Multiple and Submultiple Angle Formulas (2A, 3A, A/2)23
• Trigonometric Equations and General Solutions7
• Compound Angle Formulas (A +/- B)6
• Conditional Trigonometric Identities5
• Transformation Formulas (Product-Sum Conversions)4
• Maximum and Minimum Values of Trigonometric Expressions3
• Trigonometric Ratios of Allied Angles1

Year-wise Trigonometric Functions questions in JEE Main

Number of questions from this chapter in each year's paper(s).