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
Let $x=\sin\theta,\quad -\frac{\pi}{6}<\theta<\frac{\pi}{4}$
Then $\sqrt{1-x^2}=\cos\theta$
Expression becomes:
$\sin^{-1}\left(\frac{\sqrt{3}}{2}\sin\theta+\frac{1}{2}\cos\theta\right)$
$=\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{6}\right)\right)$
Since $\theta\in\left(-\frac{\pi}{6},\frac{\pi}{4}\right)$
$\Rightarrow \theta+\frac{\pi}{6}\in\left(0,\frac{5\pi}{12}\right)\subset\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
So principal value:
$\sin^{-1}(\sin(\theta+\frac{\pi}{6}))=\theta+\frac{\pi}{6}$
$=\sin^{-1}x+\frac{\pi}{6}$
$\boxed{\sin^{-1}x+\frac{\pi}{6}}$
Then $\sqrt{1-x^2}=\cos\theta$
Expression becomes:
$\sin^{-1}\left(\frac{\sqrt{3}}{2}\sin\theta+\frac{1}{2}\cos\theta\right)$
$=\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{6}\right)\right)$
Since $\theta\in\left(-\frac{\pi}{6},\frac{\pi}{4}\right)$
$\Rightarrow \theta+\frac{\pi}{6}\in\left(0,\frac{5\pi}{12}\right)\subset\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
So principal value:
$\sin^{-1}(\sin(\theta+\frac{\pi}{6}))=\theta+\frac{\pi}{6}$
$=\sin^{-1}x+\frac{\pi}{6}$
$\boxed{\sin^{-1}x+\frac{\pi}{6}}$
$ \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
The number of solutions of $\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$, where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, is equal to
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Solution
$\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$
$\Rightarrow \tan^{-1}\left(\frac{4x + 6x}{1 - 24x^2}\right) = \frac{\pi}{6}$
$\Rightarrow \frac{10x}{1 - 24x^2} = \frac{1}{\sqrt{3}}$
$\Rightarrow 24x^2 + 10\sqrt{3}x - 1 = 0$
$x = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48}$
$x = \frac{-10\sqrt{3} \pm \sqrt{396}}{48}$
Only one solution lies in interval $\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$
$\Rightarrow$ number of solutions $= 1$
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
Let \( \tan^{-1}x=\theta \).
Then \( \cot^{-1}x=\frac{\pi}{2}-\theta \).
Since \( x\in R \), \( -\frac{\pi}{2}<\theta<\frac{\pi}{2} \).
Given,
\( (\tan^{-1}x)^3+(\cot^{-1}x)^3=k\pi^3 \)
\( \theta^3+\left(\frac{\pi}{2}-\theta\right)^3=k\pi^3 \)
Let \( t=\frac{\theta}{\pi} \).
So, $\dfrac{-1}{2}<t<\dfrac{1}{2}$
Now,
\( k=t^3+\left(\frac{1}{2}-t\right)^3 \)
\( k=t^3+\frac{1}{8}-\frac{3t}{4}+\frac{3t^2}{2}-t^3 \)
\( k=\frac{3t^2}{2}-\frac{3t}{4}+\frac{1}{8} \)
This quadratic is minimum at \( t=\frac{1}{4} \).
\( k_{\min}=\frac{1}{32} \)
As \( t\to-\frac{1}{2} \), \( k\to\frac{7}{8} \), but \( t=-\frac{1}{2} \) is not included.
Therefore,
\( k\in\left[\frac{1}{32},\frac{7}{8}\right) \)
Answer: \( \boxed{(a)\ \left[\frac{1}{32},\frac{7}{8}\right)} \)
Then \( \cot^{-1}x=\frac{\pi}{2}-\theta \).
Since \( x\in R \), \( -\frac{\pi}{2}<\theta<\frac{\pi}{2} \).
Given,
\( (\tan^{-1}x)^3+(\cot^{-1}x)^3=k\pi^3 \)
\( \theta^3+\left(\frac{\pi}{2}-\theta\right)^3=k\pi^3 \)
Let \( t=\frac{\theta}{\pi} \).
So, $\dfrac{-1}{2}<t<\dfrac{1}{2}$
Now,
\( k=t^3+\left(\frac{1}{2}-t\right)^3 \)
\( k=t^3+\frac{1}{8}-\frac{3t}{4}+\frac{3t^2}{2}-t^3 \)
\( k=\frac{3t^2}{2}-\frac{3t}{4}+\frac{1}{8} \)
This quadratic is minimum at \( t=\frac{1}{4} \).
\( k_{\min}=\frac{1}{32} \)
As \( t\to-\frac{1}{2} \), \( k\to\frac{7}{8} \), but \( t=-\frac{1}{2} \) is not included.
Therefore,
\( k\in\left[\frac{1}{32},\frac{7}{8}\right) \)
Answer: \( \boxed{(a)\ \left[\frac{1}{32},\frac{7}{8}\right)} \)
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