My question is
$$\operatorname{arccsc}(\sqrt{37})+\operatorname{arcsin}\left(\frac{x}{\sqrt{4+x^2}}\right)=\frac{1}{2} \operatorname{arcsin}\frac{3}{5}$$
Find value of $x$.
I have converted all the functions to arctan as it's suggested to solve such kind of problem by converting to arctan, then I tried using arctan formulae but couldn't get anywhere. Could anyone help me with how to start this problem, thanks
$$\operatorname{arctan} \frac{1}{6} + \operatorname{arctan}\frac{x}{2} = \frac{1}{2} \operatorname{arcsin} \frac{3}{5}$$
Best Answer
$$\tan^{-1}\frac{1}{6}+\tan^{-1}\frac{x}{2}=a$$
Where, $a=\frac{1}{2}\sin^{-1}\frac{3}{5}$
$\sin2a=\frac{3}{5}$, further $\cos2a=\frac{4}{5}$ (you can verify using a triangle)
Now, $$\cos2a=\frac{1-\tan^2a}{1+\tan^2a}=\frac{4}{5}$$
On solving, we get $\tan a=\frac{1}{3}$
Now,
$$\tan^{-1}\frac{1}{6}+\tan^{-1}\frac{x}{2}=\tan^{-1}\frac{1}{3}$$
Using $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$
We get