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