A) \[-1\]
B) \[\frac{1}{2}\]
C) \[-\frac{1}{2}\]
D) 1
Correct Answer: B
Solution :
Let \[y={{\sin }^{2}}{{\cot }^{-1}}\left\{ \sqrt{\frac{1-x}{1+x}} \right\}\] Put \[x=\cos \theta \Rightarrow \theta ={{\cos }^{-1}}x\] Þ \[y={{\sin }^{2}}{{\cot }^{-1}}\left\{ \sqrt{\frac{1-\cos \theta }{1+\cos \theta }} \right\}={{\sin }^{2}}{{\cot }^{-1}}\left( \tan \frac{\theta }{2} \right)\] Þ \[y={{\sin }^{2}}\left( \frac{\pi }{2}-\frac{\theta }{2} \right)\] Þ \[\frac{dy}{d\theta }=2\sin \left( \frac{\pi }{2}-\frac{\theta }{2} \right)\,.\,\cos \left( \frac{\pi }{2}-\frac{\theta }{2} \right)\,\left( -\frac{1}{2} \right)\] Þ \[\frac{dy}{d\theta }=-\frac{\sin (\pi -\theta )}{2}=-\frac{\sin \theta }{2}=\frac{-1}{2}\,\sqrt{1-{{x}^{2}}}\] Þ \[\frac{dy}{dx}=\frac{dy}{d\theta }\,.\,\frac{d\theta }{dx}=\frac{-1}{2}\sqrt{1-{{x}^{2}}}\,.\,\frac{d}{dx}({{\cos }^{-1}}x)=\frac{1}{2}\].
You need to login to perform this action.
You will be redirected in
3 sec
