Jamia Millia Islamia MCA Inverse Trigonometrical Function Previous Year Questions (PYQs)
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Principal value of $\cos^{-1}(\cos 5)$ is
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Solution
For the principal range $[0, \pi]$,
$\cos^{-1}(\cos 5) = 2\pi - 5$.
If $y=\tan^{-1}\!\left(\dfrac{1+x}{1-x}\right)$, then $\dfrac{dy}{dx}$ equals …
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Solution
$y'=\dfrac{u'}{1+u^{2}}$, $u=\dfrac{1+x}{1-x}$.
$u'=\dfrac{2}{(1-x)^2}$, $1+u^2=\dfrac{2(1+x^2)}{(1-x)^2}$.
Hence $y'=\dfrac{1}{1+x^2}$.
If $y=\cos^{-1}x$ and $z=\sin^{-1}\!\sqrt{1-x^{2}}$, then $\dfrac{dy}{dz}$ equals …
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Solution
For $x\in[-1,1]$, $\sqrt{1-x^2}=\sin(\cos^{-1}x)=\sin y$.
Thus $z=\sin^{-1}(\sin y)$, so $z=y$ (up to piecewise sign); hence $\dfrac{dy}{dz}=1$.
Simplified form of $\cos^{-1}(4x^3 - 3x)$ is
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Solution
$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$
Hence, if $x = \cos\theta$, then
$\cos^{-1}(4x^3 - 3x) = \cos^{-1}(\cos 3\theta) = 3\cos^{-1}x$
$\tan^{-1}!\left(\dfrac{1}{2}\right) + \tan^{-1}!\left(\dfrac{1}{3}\right) =$
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Solution
$\tan^{-1}a + \tan^{-1}b = \tan^{-1}!\left(\dfrac{a + b}{1 - ab}\right)$
Here, $a = \dfrac{1}{2}$, $b = \dfrac{1}{3}$
$\Rightarrow \dfrac{a + b}{1 - ab} = \dfrac{\frac{5}{6}}{1 - \frac{1}{6}} = 1$
$\Rightarrow \tan^{-1}(1) = \dfrac{\pi}{4}$
$\sin(\tan^{-1}x)$, where $|x| < 1$, is equal to
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Solution
Let $\theta = \tan^{-1}x \Rightarrow \tan\theta = x$
In right triangle: opposite = $x$, adjacent = $1$
$\Rightarrow \sin\theta = \dfrac{x}{\sqrt{1 + x^2}}$
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