Evaluate: sin ^{-1}left(frac{3}{5}right) + ta | vedclass

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

Question

(A) Let $	heta = \sin ^{-1}\left(\frac{3}{5}ight)$. Then $\sin 	heta = \frac{3}{5}$.Using the identity $	an 	heta = \frac{\sin 	heta}{\sqrt{1 -

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

(A) Let $	heta = \sin ^{-1}\left(\frac{3}{5}ight)$. Then $\sin 	heta = \frac{3}{5}$.Using the identity $	an 	heta = \frac{\sin 	heta}{\sqrt{1 - \sin ^2 	heta}} = \frac{3/5}{\sqrt{1 - (3/5)^2}} = \frac{3/5}{4/5} = \frac{3}{4}$.Thus,$\sin ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$.Now,the expression becomes $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{7}ight)$.Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$ for $xy $	an ^{-1}\left(\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} 	imes \frac{1}{7}}ight) = 	an ^{-1}\left(\frac{\frac{21+4}{28}}{1 - \frac{3}{28}}ight) = 	an ^{-1}\left(\frac{25/28}{25/28}ight) = 	an ^{-1}(1) = \frac{\pi }{4}$.

Evaluate: $\sin ^{-1}\left(\frac{3}{5}\right) + \tan ^{-1}\left(\frac{1}{7}\right) = $

Similar Questions

The value of $\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

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

The value of $\tan \left( \tan^{-1} \frac{1}{2} - \tan^{-1} \frac{1}{3} \right)$ is

The real value of $x$ that satisfies the equation $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$ is

$\sin \left[ 3 \sin^{-1} \left( \frac{1}{5} \right) \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module