The value of cot left(operatorname{cosec}^{-1 | vedclass

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

Question

(B) We need to evaluate $\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+	an ^{-1} \frac{2}{3}ight)$.First,convert $\operatorname{cosec}^{-1} \fra

Vedclass · Accepted Answer

$\frac{6}{17}$

Vedclass · Answer

(B) We need to evaluate $\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+	an ^{-1} \frac{2}{3}ight)$.First,convert $\operatorname{cosec}^{-1} \frac{5}{3}$ to $	an^{-1}$. Since $\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}$,we have $\operatorname{cosec}^{-1} \frac{5}{3} = \sin^{-1} \frac{3}{5}$.Let $\sin^{-1} \frac{3}{5} = 	heta$,then $\sin 	heta = \frac{3}{5}$. Thus,$	an 	heta = \frac{3}{\sqrt{5^2-3^2}} = \frac{3}{4}$. So,$\sin^{-1} \frac{3}{5} = 	an^{-1} \frac{3}{4}$.Now the expression becomes $\cot \left(	an^{-1} \frac{3}{4} + 	an^{-1} \frac{2}{3}ight)$.Using the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left(\frac{x+y}{1-xy}ight)$,we get:$	an^{-1} \frac{3}{4} + 	an^{-1} \frac{2}{3} = 	an^{-1} \left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} 	imes \frac{2}{3}}ight) = 	an^{-1} \left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}ight) = 	an^{-1} \left(\frac{17/12}{6/12}ight) = 	an^{-1} \frac{17}{6}$.Finally,$\cot \left(	an^{-1} \frac{17}{6}ight) = \cot \left(\cot^{-1} \frac{6}{17}ight) = \frac{6}{17}$.

The value of $\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)$ is

Similar Questions

If $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$,then $x =$

If $x>0, y>0, z>0, xy+yz+zx < 1$ and if $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$, then $x+y+z$ equals to

If $2 \tan^{-1}(\cos x) = \tan^{-1}(\csc^2 x)$,then $x =$

If $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2},$ then the value of $x$ is

If $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$, then $x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module