AMU MCA Inverse Trigonometrical Function Previous Year Questions (PYQs)
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If $\theta = \sin^{-1} x + \cos^{-1} x - \tan^{-1} x, x \ge 0$ then the smallest interval in which $\theta$ lies is
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Solution
If $\cot^{-1}(\sqrt{\cos\alpha})-\tan^{-1}(\sqrt{\cos\alpha})=x$, then $\sin x$ is equal to
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Solution
Using the information from the figure, $\cos^{-1} x$ is equal to
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Solution
From the figure,
$ BC=\sqrt{AB^2-AC^2}=\sqrt{1-x^2} $
$ \Rightarrow \cos\alpha=\sqrt{1-x^2},\ \sin\alpha=x $
$ \Rightarrow \cos\beta=x,\ \sin\beta=\sqrt{1-x^2} $
$ \Rightarrow \sin^{-1}x=\alpha,\ \cos^{-1}x=\beta $
In right triangle,
$ \alpha+\beta=\frac{\pi}{2} $
$ \Rightarrow \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} $
$ \Rightarrow \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x $
$\sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right)$ is equal to
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Solution
Use identity:
$\sin^{-1}x + \sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$
$= \sin^{-1}\left(\frac{8}{17}\cdot\frac{4}{5} + \frac{3}{5}\cdot\frac{15}{17}\right)$
$= \sin^{-1}\left(\frac{32}{85} + \frac{45}{85}\right)$
$= \sin^{-1}\left(\frac{77}{85}\right)$
Also,
$\sin^{-1}x = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$
$= \tan^{-1}\left(\frac{77/85}{\sqrt{1-(77/85)^2}}\right) = \tan^{-1}\left(\frac{77}{36}\right)$
$\boxed{\text{Both (a) and (b)}}$
If $\cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$, then the value of $q$ is:
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Solution
We have, $\cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$Using the identity $\cos^{-1} \sqrt{1 - p} = \sin^{-1} \sqrt{p}$:$\cos^{-1} \sqrt{p} + \sin^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$Since $\cos^{-1} \theta + \sin^{-1} \theta = \frac{\pi}{2}$:$\frac{\pi}{2} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$$\cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4} - \frac{\pi}{2}$$\cos^{-1} \sqrt{1 - q} = \frac{\pi}{4}$$\sqrt{1 - q} = \cos\left(\frac{\pi}{4}\right)$$\sqrt{1 - q} = \frac{1}{\sqrt{2}}$Squaring both sides:$1 - q = \frac{1}{2}$$q = 1 - \frac{1}{2}$$q = \frac{1}{2}$
If $\sin^{-1} x - \cos^{-1} x = \frac{\pi}{6}$, then the value of $x$ is:
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Solution
Given, $\sin^{-1} x - \cos^{-1} x = \frac{\pi}{6}$ — (i)We know the identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ — (ii)Adding equations (i) and (ii):$(\sin^{-1} x - \cos^{-1} x) + (\sin^{-1} x + \cos^{-1} x) = \frac{\pi}{6} + \frac{\pi}{2}$$2\sin^{-1} x = \frac{\pi + 3\pi}{6}$$2\sin^{-1} x = \frac{4\pi}{6}$$2\sin^{-1} x = \frac{2\pi}{3}$$\sin^{-1} x = \frac{\pi}{3}$$x = \sin\left(\frac{\pi}{3}\right)$$x = \frac{\sqrt{3}}{2}$
The function $f(x)=\cos^2 x$ is strictly decreasing on
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Solution
$f'(x)=2\cos x(-\sin x)=-\sin 2x$
On $(0,\frac{\pi}{2})$, $\sin 2x > 0 \Rightarrow f'(x)<0$
So decreasing on $(0,\frac{\pi}{2})$
The solution set of the equation $\sin^{-1} x = 2\tan^{-1} x$ is
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Solution
Check values in domain $[-1,1]$:
$x=0,1,-1$ satisfy equation
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