Question

$ \frac{d}{dx}\left[ {{\sin }^{2}}{{\cot }^{-1}}\left\{ \sqrt{\frac{1-x}{1+x}} \right\} \right] $ equals to

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

Answer 1

Answer: (B)

Let a hyperbola, $ \frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1 $ ?(i) and its conjugate hyperbola $ -\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 $ Eccentricities of the hyperbolas (i) and (ii) are given by $ {{e}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}} $ and $ e{{}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}} $ respectively Now, $ \frac{1}{{{e}^{2}}}+\frac{1}{e{{}^{2}}}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} $ $ =\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1 $