Question

${\sin }^{-1}\frac{3}{5}-{\cos }^{-1}\frac{63}{65}={\tan }^{-1}\frac{253}{204}$

Answer 1

We have,

L.H.S.

${\sin }^{-1}\frac{3}{5}-{\cos }^{-1}\frac{63}{65}$

Let,

${\sin }^{-1}\frac{3}{5}=A$

$\sin A=\frac{3}{5}......\left(1\right)$

Using Pythagoras theorem

$H^2=P^2+B^2$

$5^2=3^2+B^2$

$B^2=25-9$

$B^2=16$

$B^2=4^2$

$B=4$

Then,

$\sin A=\frac{3}{5},$ then,

$\tan A=\frac{3}{4}$

$A={\tan }^{-1}\frac{3}{4}......\left(2\right)$

From (1) and (2) to,

${\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$

Let

${\cos }^{-1}\frac{63}{65}=\theta ......\left(3\right)$

$\cos \theta =\frac{63}{65}$

Now, using Pythagoras theorem,

$H^2=L^2+B^2$

${65}^2=L^2+{63}^2$

$4225=L^2+3969$

$4225-3969=L^2$

$L^2=256$

$L^2={16}^2$

$L=16$

Now,

$\tan \theta =\frac{16}{63}$

$\theta ={\tan }^{-1}\frac{16}{63}......\left(4\right)$

From equation (3) and (4) to,

${\cos }^{-1}\frac{63}{65}={\tan }^{-1}\frac{16}{63}$

L.H.S.

${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{16}{63}$

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

$\Rightarrow {\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{16}{63}={\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{16}{63}}{1-\frac{3}{4}\times \frac{16}{63}}\right)$

$={\tan }^{-1}\frac{\frac{3\times 63+4\times 16}{4\times 63}}{\frac{4\times 63-3\times 16}{4\times 63}}$

$={\tan }^{-1}\frac{189+64}{252-48}$

$={\tan }^{-1}\frac{253}{204}$

R.H.S.

Hence, proved.