If cot(cos^{-1} x) = sec(tan^{-1} frac{a}{sqr | vedclass

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

Question

(C) Let $\cos^{-1} x = 	heta$, then $\cos 	heta = x$. Thus, $\cot(\cos^{-1} x) = \cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{x}{\sqrt{1 -

Vedclass · Accepted Answer

$\frac{b}{\sqrt{2b^2 - a^2}}$

Vedclass · Answer

(C) Let $\cos^{-1} x = 	heta$, then $\cos 	heta = x$. Thus, $\cot(\cos^{-1} x) = \cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{x}{\sqrt{1 - x^2}}$.Let $	an^{-1} \frac{a}{\sqrt{b^2 - a^2}} = \phi$, then $	an \phi = \frac{a}{\sqrt{b^2 - a^2}}$.Using the identity $\sec^2 \phi = 1 + 	an^2 \phi$, we get $\sec^2 \phi = 1 + \frac{a^2}{b^2 - a^2} = \frac{b^2 - a^2 + a^2}{b^2 - a^2} = \frac{b^2}{b^2 - a^2}$.Therefore, $\sec \phi = \frac{b}{\sqrt{b^2 - a^2}}$.Equating the two sides: $\frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}}$.Squaring both sides: $\frac{x^2}{1 - x^2} = \frac{b^2}{b^2 - a^2}$.$x^2(b^2 - a^2) = b^2(1 - x^2) = b^2 - b^2 x^2$.$x^2(b^2 - a^2 + b^2) = b^2 \implies x^2(2b^2 - a^2) = b^2$.$x^2 = \frac{b^2}{2b^2 - a^2} \implies x = \frac{b}{\sqrt{2b^2 - a^2}}$.

If $\cot(\cos^{-1} x) = \sec(\tan^{-1} \frac{a}{\sqrt{b^2 - a^2}})$, then the value of $x$ is

Similar Questions

If $\cot ^{-1}(\alpha)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots$ up to $100$ terms,then $\alpha$ is

The value of $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$ (where,$x, y>0$) is

If $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$, then $5+x=$

Prove $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}$,where $x \in\left(0, \frac{\pi}{4}\right)$.

Prove that $2 \sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{24}{7}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module