If {sin ^{ - 1}}left( {frac{{2a}}{{1 + {a^2}} | vedclass

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

Question

(D) Given the equation: ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} ight) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} ight) = 2{	an ^{

Vedclass · Accepted Answer

$\frac{{a + b}}{{1 - ab}}$

Vedclass · Answer

(D) Given the equation: ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} ight) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} ight) = 2{	an ^{ - 1}}x$Substitute $a = 	an 	heta$ and $b = 	an \phi$:${\sin ^{ - 1}}\left( {\frac{{2	an 	heta }}{{1 + {{	an }^2}	heta }}} ight) + {\sin ^{ - 1}}\left( {\frac{{2	an \phi }}{{1 + {{	an }^2}\phi }}} ight) = 2{	an ^{ - 1}}x$Using the identity $\sin 2	heta = \frac{2	an 	heta}{1 + 	an^2 	heta}$,we get:${\sin ^{ - 1}}(\sin 2	heta) + {\sin ^{ - 1}}(\sin 2\phi) = 2{	an ^{ - 1}}x$Simplifying the expression:$2	heta + 2\phi = 2{	an ^{ - 1}}x$Dividing by $2$:$	heta + \phi = {	an ^{ - 1}}x$Taking the tangent of both sides:$x = 	an(	heta + \phi)$Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$:$x = \frac{	an 	heta + 	an \phi}{1 - 	an 	heta 	an \phi}$Substituting $a$ and $b$ back:$x = \frac{a + b}{1 - ab}$

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

Similar Questions

Find the value of $\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$

If $4 \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$,then $x =$ . . . . . . .

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ is equal to

If $x>0, y>0, z>0, xy+yz+zx < 1$ and if $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$, then $x+y+z$ equals to

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then the value of $x^{2025}+x^{2026}+x^{2027}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module