Question

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

• A. $ 2\sin 2x $
• B. $ \sin 2x $
• C. $ \frac{1}{2} $
• D. $ -\frac{1}{2} $

Answer 1

The Correct Option is D

$ y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+x}{1-x}} $ Put, $ x=\cos \theta $
$ \Rightarrow $ $ \theta ={{\cos }^{-1}}x $ ...(i) $ y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }} $ $ y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{2{{\cos }^{2}}\theta /2}{2{{\sin }^{2}}\theta /2}} $ $ y={{\sin }^{2}}{{\cot }^{-1}}(\cot \theta /2) $ $ y={{\sin }^{2}}\theta /2 $ $ y=\frac{1-\cos \theta }{2} $ $ y=\frac{1-x}{2} $ [From E(i)] Differentiating w.r.t. $ x, $ we get $ \frac{dy}{dx}=-\frac{1}{2} $