If y = sin^{2} (cot^{-1} sqrt{frac{1 + x}{1 - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation is given as,$y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$.Substitute $x = \cos 2	heta$,then $\frac{1 + x}{1 - x} = \frac{1 +

Vedclass · Accepted Answer

$\frac{- 1}{2}$

Vedclass · Answer

(A) The equation is given as,$y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$.Substitute $x = \cos 2	heta$,then $\frac{1 + x}{1 - x} = \frac{1 + \cos 2	heta}{1 - \cos 2	heta} = \frac{2\cos^{2}	heta}{2\sin^{2}	heta} = \cot^{2}	heta$.So,$y = \sin^{2} (\cot^{-1} \sqrt{\cot^{2}	heta}) = \sin^{2} (\cot^{-1} (\cot	heta)) = \sin^{2}	heta$.Using the identity $\sin^{2}	heta = \frac{1 - \cos 2	heta}{2}$,we get $y = \frac{1 - x}{2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (\frac{1}{2} - \frac{x}{2}) = 0 - \frac{1}{2} = -\frac{1}{2}$.

If $y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$,then $\frac{dy}{dx} = $

Similar Questions

If $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=k$ then $\sin ^{-1}\left(\sqrt{\frac{k}{2}}\right)+\cos ^{-1}\left(\frac{k}{3}\right)=$

The sum $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

If $\tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8}$, then $x^2=$

If $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$,then $\operatorname{cosec}^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C=$

Show that $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$

Vedclass Test Series

Exam Paper Generator

Online Exam Module