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

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

Question

(D) Let the given expression be $S = \sin \left(	an ^{-1} \frac{4}{5}+	an ^{-1} \frac{4}{3}+	an ^{-1} \frac{1}{9}-	an ^{-1} \frac{1}{7}ight)$.Re

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Let the given expression be $S = \sin \left(	an ^{-1} \frac{4}{5}+	an ^{-1} \frac{4}{3}+	an ^{-1} \frac{1}{9}-	an ^{-1} \frac{1}{7}ight)$.Rearranging the terms, we get $S = \sin \left(\left(	an ^{-1} \frac{4}{5}-	an ^{-1} \frac{1}{7}ight) + \left(	an ^{-1} \frac{4}{3}+	an ^{-1} \frac{1}{9}ight)ight)$.Let $p = 	an ^{-1} \frac{4}{5}-	an ^{-1} \frac{1}{7}$. Using the formula $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \left(\frac{x-y}{1+xy}ight)$, we have $p = 	an ^{-1} \left(\frac{\frac{4}{5}-\frac{1}{7}}{1+\frac{4}{5} \cdot \frac{1}{7}}ight) = 	an ^{-1} \left(\frac{\frac{28-5}{35}}{\frac{35+4}{35}}ight) = 	an ^{-1} \left(\frac{23}{39}ight)$.Let $q = 	an ^{-1} \frac{4}{3}+	an ^{-1} \frac{1}{9}$. Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left(\frac{x+y}{1-xy}ight)$, we have $q = 	an ^{-1} \left(\frac{\frac{4}{3}+\frac{1}{9}}{1-\frac{4}{3} \cdot \frac{1}{9}}ight) = 	an ^{-1} \left(\frac{\frac{12+1}{9}}{\frac{27-4}{27}}ight) = 	an ^{-1} \left(\frac{13}{9} \cdot \frac{27}{23}ight) = 	an ^{-1} \left(\frac{39}{23}ight)$.Since $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$, we have $q = \cot ^{-1} \left(\frac{23}{39}ight) = \frac{\pi}{2} - 	an ^{-1} \left(\frac{23}{39}ight)$.Thus, $p+q = 	an ^{-1} \left(\frac{23}{39}ight) + \frac{\pi}{2} - 	an ^{-1} \left(\frac{23}{39}ight) = \frac{\pi}{2}$.Therefore, $S = \sin \left(\frac{\pi}{2}ight) = 1$.

$\sin \left(\tan ^{-1} \frac{4}{5}+\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}-\tan ^{-1} \frac{1}{7}\right) = $

Similar Questions

Find $\frac{dy}{dx}$ if $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

If $x = \sin (2 \tan^{-1} 2)$,$y = \cos (2 \tan^{-1} 3)$,and $z = \sec (2 \tan^{-1} 4)$,then:

The number of real solutions of $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$ is

If $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$, then $x \in$

$\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module