Difficulty distribution
How the classified questions are distributed by difficulty.
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Practice Inverse Trigonometric Functions - Trigonometry - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.
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Year-wise coverage for Inverse Trigonometric Functions. Each bar uses a separate theme-derived color.
How the classified questions are distributed by difficulty.
MCQ, numerical, multiple-select and other formats found in these papers.
Top subjects by unique question coverage.
Top topics across the included previous year papers.
Top subtopics inside this exact selection.
Question coverage for the most populated papers. Every active PYP paper remains listed below.
Newest papers appear first. Sort by year, question coverage or name.
| Paper | Year / session | Questions in this view | Open |
|---|---|---|---|
| WB JEE 2026 | 2026 | 5 | View paper |
| WB JEE 2026 | 2026 | 5 | View paper |
| VITEEE 2025 | 2025 | 1 | View paper |
| WB JEE 2025 | 2025 | 2 | View paper |
| WB JEE 2021 | 2021 | 1 | View paper |
| WB JEE 2018 | 2018 | 1 | View paper |
| WB JEE 2017 | 2017 | 1 | View paper |
| WB JEE 2016 | 2016 | 1 | View paper |
| WB JEE 2011 | 2011 | 2 | View paper |
| WB JEE 2010 | 2010 | 1 | View paper |
| WB JEE 2009 | 2009 | 1 | View paper |
| WB JEE 2008 | 2008 | 1 | View paper |
Practice every matching question in batches of 20, with every available option.
If $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$, then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ is equal to
The number of solutions of $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$ is
$S=\cot ^{-1}\left(\frac{1+2 \times 6}{4}\right)+\cot ^{-1}\left(\frac{1+3 \times 8}{10}\right)+\cot ^{-1} \left(\frac{1+4 \times 10}{18}\right) \ldots \ldots \ldots$ upto $\infty$ terms is equal to
The value of \(\tan \alpha + 2\tan (2\alpha ) + 4\tan (4\alpha ) + ... + {2^{n - 1}}\tan ({2^{n - 1}}\alpha ) + {2^n}\cot ({2^n}\alpha )\) is
\(\tan \left[ {{\pi \over 4} + {1 \over 2}{{\cos }^{ - 1}}\left( {{a \over b}} \right)} \right] + \tan \left[ {{\pi \over 4} - {1 \over 2}{{\cos }^{ - 1}}\left( {{a \over b}} \right)} \right]\) is equal to
Value of \({\tan ^{ - 1}}\left( {{{\sin 2 - 1} \over {\cos 2}}} \right)\) is
If \({r^2} = {x^2} + {y^2} + {z^2}\), then prove that \({\tan ^{ - 1}}\left( {{{yz} \over {rx}}} \right) + {\tan ^{ - 1}}\left( {{{zx} \over {ry}}} \right) + {\tan ^{ - 1}}\left( {{{xy} \over {rz}}} \right) = {\pi \over 2}\)
The solution set of the inequation \({\cos ^{ - 1}}x < {\sin ^{ - 1}}x\) is
If for two real numbers $\mathrm{a}, \mathrm{b}$ with $|\mathrm{a}| \leq 1$ and $|\mathrm{b}| \leq 1$,
$\frac{1}{3}+\frac{\sin ^{-1} a+\sin ^{-1} b}{4}+\frac{\left(\sin ^{-1} a+\sin ^{-1} b\right)^2}{16}+\frac{\left(\sin ^{-1} a+\sin ^{-1} b\right)^3}{64}+\cdots=\frac{2(8-3 \pi)}{3(16+3 \pi)}, \quad$ then the value of $\sin ^{-1}\left(a \sqrt{1-b^2}+b \sqrt{1-a^2}\right)$ is
The true set of values of ' $K$ ' for which $\sin ^{-1}\left(\frac{1}{1+\sin ^2 x}\right)=\frac{K \pi}{6}$ may have a solution is
If $\sum\limits_{r=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 r^2}\right)=a$, then $\tan a$ is equal to
If $f(x)=x\left(1331 x^2-3630 x+3300\right)$, then for $a=\cos ^2\left(\tan ^{-1}\left(\sin \left(\cot ^{-1} 3\right)\right)\right)$
Let $g(x)=a x+b$, where $a<0$ and $g$ is defined from $[1,3]$ onto $[0,2]$. Then the value of $\cot \left(\cos ^{-1}(|\sin x|+|\cos x|)+\right. \left.\sin ^{-1}(-|\cos x|-|\sin x|)\right)$ is equal to