cos ^{ - 1}frac{4}{5} + tan ^{ - 1}frac{3}{5} | vedclass

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

Question

(A) Let $x = \cos ^{ - 1}\frac{4}{5}$. Then $\cos x = \frac{4}{5}$.Since $\cos x = \frac{	ext{adj}}{	ext{hyp}} = \frac{4}{5}$,the opposite side is $

Vedclass · Accepted Answer

$	an ^{ - 1}\frac{27}{11}$

Vedclass · Answer

(A) Let $x = \cos ^{ - 1}\frac{4}{5}$. Then $\cos x = \frac{4}{5}$.Since $\cos x = \frac{	ext{adj}}{	ext{hyp}} = \frac{4}{5}$,the opposite side is $\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$.Thus,$	an x = \frac{	ext{opp}}{	ext{adj}} = \frac{3}{4}$,which implies $x = 	an ^{ - 1}\frac{3}{4}$.Now,the expression becomes $	an ^{ - 1}\frac{3}{4} + 	an ^{ - 1}\frac{3}{5}$.Using the formula $	an ^{ - 1}A + 	an ^{ - 1}B = 	an ^{ - 1}\left( \frac{A + B}{1 - AB} ight)$:$= 	an ^{ - 1}\left( \frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{3}{4} 	imes \frac{3}{5}} ight)$$= 	an ^{ - 1}\left( \frac{\frac{15 + 12}{20}}{1 - \frac{9}{20}} ight)$$= 	an ^{ - 1}\left( \frac{\frac{27}{20}}{\frac{11}{20}} ight)$$= 	an ^{ - 1}\frac{27}{11}$.

$\cos ^{ - 1}\frac{4}{5} + \tan ^{ - 1}\frac{3}{5} = $

Similar Questions

Find $\frac{dy}{dx}$ for $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ where $|x| < 1$.

The value of $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ is

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then

Evaluate $\sin(3 \sin^{-1}(1/5))$.

If $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$,then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module