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, $\frac{1}{e^2}+\frac{1}{e{}^2}=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}$ $=\frac{a^2+b^2}{a^2+b^2}=1$