sec(text{cosec}^{-1}x) is equal to | vedclass

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

Question

(A) Let $	heta = 	ext{cosec}^{-1}x$. Then $	ext{cosec}	heta = x$,where $|x| \ge 1$ and $	heta \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$.We

Vedclass · Accepted Answer

$	ext{cosec}(\sec^{-1}x)$

Vedclass · Answer

(A) Let $	heta = 	ext{cosec}^{-1}x$. Then $	ext{cosec}	heta = x$,where $|x| \ge 1$ and $	heta \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$.We need to find $\sec(	ext{cosec}^{-1}x) = \sec	heta$.Since $	ext{cosec}	heta = x$,we have $\sin	heta = \frac{1}{x}$.Using the identity $\cos^2	heta = 1 - \sin^2	heta = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$,we get $\cos	heta = \frac{\sqrt{x^2 - 1}}{|x|}$.Therefore,$\sec	heta = \frac{1}{\cos	heta} = \frac{|x|}{\sqrt{x^2 - 1}}$.Now,consider $	ext{cosec}(\sec^{-1}x)$. Let $\phi = \sec^{-1}x$. Then $\sec\phi = x$,so $\cos\phi = \frac{1}{x}$.Then $\sin\phi = \sqrt{1 - \cos^2\phi} = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|}$.Thus,$	ext{cosec}\phi = \frac{1}{\sin\phi} = \frac{|x|}{\sqrt{x^2 - 1}}$.Since both expressions equal $\frac{|x|}{\sqrt{x^2 - 1}}$,we have $\sec(	ext{cosec}^{-1}x) = 	ext{cosec}(\sec^{-1}x)$.

$\sec(\text{cosec}^{-1}x)$ is equal to

Similar Questions

Let $S_{k} = \sum_{r=1}^{k} \tan^{-1}\left(\frac{6^{r}}{2^{2r+1} + 3^{2r+1}}\right)$. Then $\lim_{k \rightarrow \infty} S_{k}$ is equal to

$2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = $

$2{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{\theta }{2}} \right] = $

If $\cos^{-1} x = \alpha$ $(0 < x < 1)$ and $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$,then $\alpha$ is

If $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$, then $x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module