If a b 0 and sec^{-1} left( frac{a+b}{a-b | vedclass

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

Question

(B) Given $\sec^{-1} \left( \frac{a+b}{a-b} \right) = 2 \sin^{-1} x$. Taking the reciprocal,we get $\cos^{-1} \left( \frac{a-b}{a+b} \right) = 2 \sin^

Vedclass · Accepted Answer

$\sqrt{\frac{b}{a+b}}$

Vedclass · Answer

(B) Given $\sec^{-1} \left( \frac{a+b}{a-b} ight) = 2 \sin^{-1} x$. Taking the reciprocal,we get $\cos^{-1} \left( \frac{a-b}{a+b} ight) = 2 \sin^{-1} x$. Dividing the numerator and denominator by $a$,we have $\cos^{-1} \left( \frac{1 - b/a}{1 + b/a} ight) = 2 \sin^{-1} x$. Using the identity $\cos^{-1} \left( \frac{1 - y^2}{1 + y^2} ight) = 2 	an^{-1} y$,where $y = \sqrt{b/a}$,we get $2 	an^{-1} \left( \sqrt{\frac{b}{a}} ight) = 2 \sin^{-1} x$. Dividing by $2$,we have $	an^{-1} \left( \sqrt{\frac{b}{a}} ight) = \sin^{-1} x$. Using the identity $	an^{-1} y = \sin^{-1} \left( \frac{y}{\sqrt{1+y^2}} ight)$,we get $\sin^{-1} \left( \frac{\sqrt{b/a}}{\sqrt{1 + b/a}} ight) = \sin^{-1} x$. Simplifying the expression inside,we get $\sin^{-1} \left( \sqrt{\frac{b/a}{(a+b)/a}} ight) = \sin^{-1} x$. Thus,$x = \sqrt{\frac{b}{a+b}}$.

If $a > b > 0$ and $\sec^{-1} \left( \frac{a+b}{a-b} \right) = 2 \sin^{-1} x$,then $x$ is

Similar Questions

The value of $\operatorname{cosec}^{-1}(\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)-\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is equal to

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$,where $-1 \leq x \leq 1$,$-3 \leq y \leq 3$,and $x \leq \frac{y}{3}$,then for all $x, y$,$9x^2 - 6xy \cos \alpha + y^2$ is equal to

Find the value of $\tan \frac{1}{2} \left[ \sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2} \right]$,where $|x| <1, y>0$ and $xy < 1$.

Evaluate: $\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)$

$A$ possible value of $\tan \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module