begin{aligned} & text{If } cot left(cos ^{-1} | vedclass

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

Question

(A) Given that,$\cot \left(\cos ^{-1} xight)=\sec \left\{	an ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}ight)ight\}$.Since $\cos ^{-1} x = \cot ^{-1} \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given that,$\cot \left(\cos ^{-1} xight)=\sec \left\{	an ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}ight)ight\}$.Since $\cos ^{-1} x = \cot ^{-1} \left(\frac{x}{\sqrt{1-x^2}}ight)$ and $	an ^{-1} 	heta = \sec ^{-1} \left(\sqrt{1+	heta^2}ight)$,we have:$\cot \left(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}ight) = \sec \left\{\sec ^{-1} \sqrt{1+\left(\frac{a}{\sqrt{b^2-a^2}}ight)^2}ight\}$$\Rightarrow \frac{x}{\sqrt{1-x^2}} = \sqrt{1+\frac{a^2}{b^2-a^2}} = \sqrt{\frac{b^2-a^2+a^2}{b^2-a^2}} = \frac{b}{\sqrt{b^2-a^2}}$.Squaring both sides,we get:$\frac{x^2}{1-x^2} = \frac{b^2}{b^2-a^2}$$x^2(b^2-a^2) = b^2(1-x^2)$$x^2 b^2 - x^2 a^2 = b^2 - x^2 b^2$$2 x^2 b^2 - x^2 a^2 = b^2$$x^2(2 b^2 - a^2) = b^2$$x^2 = \frac{b^2}{2 b^2 - a^2}$$x = \frac{b}{\sqrt{2 b^2 - a^2}}$.

$\begin{aligned} & \text{If } \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\} \\ & b>a, \text{ then } x= \end{aligned}$

Similar Questions

$\sin^{-1} x + \cos^{-1} x$ is equal to

If $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$, then $1-x y-y z-z x$ is equal to

The value of $\sin^{-1} \left( \frac{12}{13} \right) - \sin^{-1} \left( \frac{3}{5} \right)$ is equal to

$\operatorname{Tan}^{-1} \left( \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1} \right) + \operatorname{Tan}^{-1} \left( \frac{1}{\sqrt{5}} \right) =$

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f(x) = \tan^{-1}(\sin x + \cos x)$ in the interval $[0, \frac{\pi}{2}]$. Then the value of $\tan(M - m)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module