Find the value of $x$ from the following equation in terms of $a$ and $b$ $$\sin^{-1} {2a\over{1+a^2}} + \sin^{-1}{2b\over{1+b^2}} = 2\tan^{-1}x$$
I tried to expand the LHS using the formula $$\sin^{-1}c+\sin^{-1}d = \sin^{-1}\left(c\sqrt{1+d^2} + d\sqrt{1+c^2}\right)$$
But it didn't work out. Could anyone help me out?
Divide both sides by $2$, then using the $\arctan(x)=y \implies x=\tan(y)$ you will get that $$x=\tan\left(\frac{\arcsin\left(\frac{2a}{1+a^2}\right)+\arcsin\left(\frac{2b}{1+b^2}\right)}{2}\right)$$
\arctan and \tan instead of tan and arctan.
– saulspatz
Sep 08 '18 at 13:02
$2$ comes out as $2$.
– saulspatz
Sep 08 '18 at 13:09
\left( ... \right) instead of just ( ... ) for parentheses that need to contain large expressions; they will grow as needed.
– Théophile
Sep 08 '18 at 13:11
Let $\arctan a=u,a=\tan u$
If $-\dfrac\pi2\le2u\le\dfrac\pi2\iff -1\le a\le1$ $$P=\arcsin\dfrac{2a}{1+a^2}=2\arctan a$$
If $2u>\dfrac\pi2,P=\pi-2\arctan a$
If $2u<-\dfrac\pi2,P=-\pi-2\arctan a$
Now use my answer in showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
