2 pi - left(sin ^{-1} frac{4}{5} + sin ^{-1} | vedclass

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

Question

(D) Let $x = \sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}$.First,convert $\sin ^{-1} \frac{4}{5}$ and $\sin ^{-1} \frac

Vedclass · Accepted Answer

$\frac{3 \pi}{2}$

Vedclass · Answer

(D) Let $x = \sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}$.First,convert $\sin ^{-1} \frac{4}{5}$ and $\sin ^{-1} \frac{5}{13}$ to $	an ^{-1}$ form:$\sin ^{-1} \frac{4}{5} = 	an ^{-1} \frac{4}{3}$ and $\sin ^{-1} \frac{5}{13} = 	an ^{-1} \frac{5}{12}$.Now,calculate the sum $	an ^{-1} \frac{4}{3} + 	an ^{-1} \frac{5}{12}$:$	an ^{-1} \left(\frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3} 	imes \frac{5}{12}}ight) = 	an ^{-1} \left(\frac{\frac{16+5}{12}}{\frac{36-20}{36}}ight) = 	an ^{-1} \left(\frac{21}{12} 	imes \frac{36}{16}ight) = 	an ^{-1} \left(\frac{63}{16}ight)$.Since $	an ^{-1} \frac{63}{16} = \cot ^{-1} \frac{16}{63}$,we convert this to $\sin ^{-1}$ form. Let $	heta = 	an ^{-1} \frac{63}{16}$,then $	an 	heta = \frac{63}{16}$,so $\sin 	heta = \frac{63}{65}$.Thus,$\sin ^{-1} \frac{63}{65} + \sin ^{-1} \frac{16}{65} = \frac{\pi}{2}$ (using $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$ and $\sin ^{-1} \frac{16}{65} = \cos ^{-1} \frac{63}{65}$).The expression becomes $2 \pi - \frac{\pi}{2} = \frac{3 \pi}{2}$.

$2 \pi - \left(\sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}\right)$ is equal to

Similar Questions

The equation of the line touching both the parabolas $y^{2}=4x$ and $x^{2}=-32y$ is

$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x=$

The electronic configuration of an element is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$. This represents its

$C_6H_6 + 3O_3$ $\longrightarrow X \stackrel{Zn / H_2O}{}$ ${\longrightarrow} Y$; $X$ and $Y$ are respectively

Which of the following complexes has a square planar geometry?

Vedclass Test Series

Exam Paper Generator

Online Exam Module