cos ^{-1} frac{3}{5} + sin ^{-1} frac{5}{13} | vedclass

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

Question

(A) We know that $\cos ^{-1} \left(\frac{3}{5}ight) = 	an ^{-1} \left(\frac{4}{3}ight)$ and $\sin ^{-1} \left(\frac{5}{13}ight) = 	an ^{-1} \l

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) We know that $\cos ^{-1} \left(\frac{3}{5}ight) = 	an ^{-1} \left(\frac{4}{3}ight)$ and $\sin ^{-1} \left(\frac{5}{13}ight) = 	an ^{-1} \left(\frac{5}{12}ight)$.Substituting these into the expression:$	an ^{-1} \left(\frac{4}{3}ight) + 	an ^{-1} \left(\frac{5}{12}ight) + 	an ^{-1} \left(\frac{16}{63}ight)$Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left(\frac{x+y}{1-xy}ight)$ for the first two terms:$= 	an ^{-1} \left[\frac{\frac{4}{3} + \frac{5}{12}}{1 - \left(\frac{4}{3} 	imes \frac{5}{12}ight)}ight] + 	an ^{-1} \left(\frac{16}{63}ight)$$= 	an ^{-1} \left[\frac{\frac{16+5}{12}}{\frac{36-20}{36}}ight] + 	an ^{-1} \left(\frac{16}{63}ight) = 	an ^{-1} \left(\frac{21}{12} 	imes \frac{36}{16}ight) + 	an ^{-1} \left(\frac{16}{63}ight)$$= 	an ^{-1} \left(\frac{63}{16}ight) + 	an ^{-1} \left(\frac{16}{63}ight)$Since $	an ^{-1} x + 	an ^{-1} \left(\frac{1}{x}ight) = \frac{\pi}{2}$ for $x > 0$, we get:$= \frac{\pi}{2}$.

$\cos ^{-1} \frac{3}{5} + \sin ^{-1} \frac{5}{13} + \tan ^{-1} \frac{16}{63} = $

Similar Questions

Assume that $3.13 \leq \pi \leq 3.15$. The integer closest to the value of $\sin ^{-1}(\sin 1 \cos 4+\cos 1 \sin 4)$,where $1$ and $4$ appearing in $\sin$ and $\cos$ are given in radians,is

If $\tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$,then $x =$

Let $f(x) = \cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}\right)\right)$ for $0 < x < 1$. Then :

$\left[\sin \left(\tan ^{-1} \frac{3}{4}\right)\right]^{2}+\left[\sin \left(\tan ^{-1} \frac{4}{3}\right)\right]^{2}=$

If $y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module