Question

Prove $2sin^{-1}\ \frac {3}{5}=tan^{-1}\ \frac {24}{7}$

Answer 1

Let sin-1$\frac 35$ = x. Then, sinx = $\frac 35$
$\implies$cosx = $\sqrt {1- (\frac 35)^2}$ = $\frac 45$,
therefore tanx = $\frac 34$
therefore x = tan-1$\frac 34$
$\implies$sin-1$\frac 35$=tan-1$\frac 34$
Now, we have:
LHS = 2sin-1$\frac 35$
= 2tan-1$\frac 34$
=tan-1$\frac {2 \times \frac 34}{1-(\frac 34)^2}$ [2 tan-1x = tan-1$\frac {2x}{1-x^2}$]
= tan-1($\frac 32 \times \frac {16}{7}$)
= tan-1$\frac {24}{7}$
= RHS