sin ^{-1} frac{4}{5} + 2 tan ^{-1} frac{1}{3} | vedclass

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

Question

(C) We are given the expression: $\sin ^{-1} \frac{4}{5} + 2 	an ^{-1} \frac{1}{3}$.First,use the formula $2 	an ^{-1} x = 	an ^{-1} \frac{2x}{1-x^

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(C) We are given the expression: $\sin ^{-1} \frac{4}{5} + 2 	an ^{-1} \frac{1}{3}$.First,use the formula $2 	an ^{-1} x = 	an ^{-1} \frac{2x}{1-x^2}$:$2 	an ^{-1} \frac{1}{3} = 	an ^{-1} \frac{2(1/3)}{1-(1/3)^2} = 	an ^{-1} \frac{2/3}{1-1/9} = 	an ^{-1} \frac{2/3}{8/9} = 	an ^{-1} \left( \frac{2}{3} 	imes \frac{9}{8} ight) = 	an ^{-1} \frac{3}{4}$.Now,the expression becomes $\sin ^{-1} \frac{4}{5} + 	an ^{-1} \frac{3}{4}$.Since $	an ^{-1} \frac{3}{4} = 	heta$,then $	an 	heta = \frac{3}{4}$,which implies $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.Thus,$	an ^{-1} \frac{3}{4} = \cos ^{-1} \frac{4}{5}$.Substituting this back,we get $\sin ^{-1} \frac{4}{5} + \cos ^{-1} \frac{4}{5}$.Using the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we get $\frac{\pi}{2}$.

$\sin ^{-1} \frac{4}{5} + 2 \tan ^{-1} \frac{1}{3}$ is equal to

Similar Questions

If $x^2+\alpha y^2+2 \beta y=a^2$ represents a pair of perpendicular lines,then $\beta$ equals to

The element used by Rutherford in his famous scattering experiment was

The melting point of $Zn$ is lower as compared to those of the other elements of the $3d$ series because:

If the mean of a Poisson distribution is $\frac{1}{2}$,then the ratio of $P(X=3)$ to $P(X=2)$ is

Two vibrating tuning forks produce progressive waves given by $Y_1 = 4 \sin(500\pi t)$ and $Y_2 = 2 \sin(506\pi t)$. The number of beats produced per minute is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module