{sin ^{ - 1}}frac{4}{5} + 2{tan ^{ - 1}}frac{ | vedclass

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Question

(A) We are given the expression: ${\sin ^{ - 1}}\frac{4}{5} + 2{	an ^{ - 1}}\frac{1}{3}$.First,convert ${\sin ^{ - 1}}\frac{4}{5}$ into the $	an^{-1

Vedclass · Accepted Answer

$\frac{\pi }{2}$

Vedclass · Answer

(A) We are given the expression: ${\sin ^{ - 1}}\frac{4}{5} + 2{	an ^{ - 1}}\frac{1}{3}$.First,convert ${\sin ^{ - 1}}\frac{4}{5}$ into the $	an^{-1}$ form. Let ${\sin ^{ - 1}}\frac{4}{5} = 	heta$,then $\sin 	heta = \frac{4}{5}$.Using the identity $	an 	heta = \frac{\sin 	heta}{\sqrt{1 - \sin^2 	heta}} = \frac{4/5}{\sqrt{1 - 16/25}} = \frac{4/5}{3/5} = \frac{4}{3}$.So,${\sin ^{ - 1}}\frac{4}{5} = {	an ^{ - 1}}\frac{4}{3}$.Next,use the formula $2{	an ^{ - 1}}x = {	an ^{ - 1}}\frac{2x}{1 - x^2}$ for $2{	an ^{ - 1}}\frac{1}{3}$:$2{	an ^{ - 1}}\frac{1}{3} = {	an ^{ - 1}}\left( \frac{2(1/3)}{1 - (1/3)^2} ight) = {	an ^{ - 1}}\left( \frac{2/3}{1 - 1/9} ight) = {	an ^{ - 1}}\left( \frac{2/3}{8/9} ight) = {	an ^{ - 1}}\left( \frac{2}{3} 	imes \frac{9}{8} ight) = {	an ^{ - 1}}\frac{3}{4}$.Now,the expression becomes ${	an ^{ - 1}}\frac{4}{3} + {	an ^{ - 1}}\frac{3}{4}$.Since ${	an ^{ - 1}}x + {	an ^{ - 1}}\frac{1}{x} = \frac{\pi }{2}$ for $x > 0$,we have ${	an ^{ - 1}}\frac{4}{3} + {	an ^{ - 1}}\frac{3}{4} = \frac{\pi }{2}$.Thus,the correct option is $A$.

${\sin ^{ - 1}}\frac{4}{5} + 2{\tan ^{ - 1}}\frac{1}{3} = $

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