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

Two identical capacitors $A$ and $B$ are charged to the same potential $V$ and are connected in two circuits at $t = 0$,as shown in the figure. The charge on the capacitors at time $t = CR$ are respectively:

Let $z = a - \frac{i}{2}$,where $a \in R$. Then $|i + z|^2 - |i - z|^2$ is equal to

The area (in square units) of the region bounded by the curve $y=|\sin 2x|$ and the $X$-axis in the interval $[0, 2\pi]$ is:

The number of lone pairs of electrons present on $Xe$ in $XeF_4$ is

Golden rice is a transgenic crop of the future with the following improved trait:

Vedclass Test Series

Exam Paper Generator

Online Exam Module