JEE MAIN Inverse Trigonometrical Function Previous Year Questions (PYQs) – Page 1 of 7
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$\tan\!\left(2\tan^{-1}\!\tfrac{1}{5}+\sec^{-1}\!\tfrac{\sqrt{5}}{2}+2\tan^{-1}\!\tfrac{1}{8}\right)$ is equal to:
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Solution
Considering the principal values of the inverse trigonometric functions, $\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right),-\frac{1}{2}< x<\frac{1}{\sqrt{2}}$, is equal to
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Solution
$ \text{If } 0 < x < \tfrac{1}{\sqrt{2}} \text{ and } \tfrac{\sin^{-1}x}{\alpha} = \tfrac{\cos^{-1}x}{\beta}, \text{ then the value of } \sin!\left(\tfrac{2\pi\alpha}{\alpha+\beta}\right) \text{ is :}$
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Solution
The value of $\cot\!\left(\displaystyle\sum_{n=1}^{19}\cot^{-1}\!\left(1+\sum_{p=1}^{n}2p\right)\right)$ is :
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Solution
The sum of the infinite series $\cot^{-1}\left(\dfrac{7}{4}\right)+\cot^{-1}\left(\dfrac{19}{4}\right)+\cot^{-1}\left(\dfrac{30}{4}\right)+\cot^{-1}\left(\dfrac{67}{4}\right)+\cdots$ is:
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Solution
Let $S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$.
If $\mathrm{n(S)}$ denotes the number of elements in $\mathrm{S}$ then :
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Solution
The domain of the function $f(x)=\sin^{-1}!\big([,2x^{2}-3,]\big)+\log_{2}!\left(\log_{1/2}(x^{2}-5x+5)\right)$, where $[,\cdot,]$ is the greatest integer function, is:
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Solution
A possible value of $\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$ is :
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Solution
The set of all values of k for which ${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3},\,x \in R$, is the interval :
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Solution
Considering only the principal values of the inverse trigonometric functions, the domain of the function
$f(x)=\cos^{-1}!\left(\dfrac{x^{2}-4x+2}{x^{2}+3}\right)$ is:
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Solution
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