#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.13 Question 5  Maths Textbook Solution.

Given:
$\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}$
Hint:
First, we convert $\cos^{-1}$ to $\sin^{-1}$ and then use $\sin \left ( a+b \right )$ formula.
Solution:
Let          $a= \cos^{-1}\frac{12}{13}$ and $b= \sin^{-1}\frac{3}{5}$
Finding $\sin a , \cos a$.
Let          $a= \cos^{-1}\frac{12}{13}$
$\cos a= \left (\frac{12}{13} \right )$
We know that
$\sin a= \sqrt{1-\cos ^{2}}a$
$= \sqrt{1-\left ( \frac{12}{13} \right )^{2}}$
$= \sqrt{\frac{25}{169}}$
$= \frac{5}{13}$
Again, finding $\sin b, \cos b$
Let          $b= \sin^{-1}\left ( \frac{3}{5} \right )$
$\Rightarrow \sin b= \frac{3}{5}$
We know that,
$\cos b= \sqrt{1-\sin ^{2}b}$
$\Rightarrow \cos b= \sqrt{1-\left ( \frac{3}{5} \right )^{2}}$
$= \sqrt{\frac{16}{25}}= \frac{4}{5}$
$\cos b= \frac{4}{5}$
We know that,
$\sin \left ( a+b \right )= \sin a\cos b+\cos a\sin b$
Putting, $\sin a= \frac{5}{13},\cos a= \frac{12}{13}$
$\sin b= \frac{3}{5},\cos b= \frac{4}{5}$
We have,
$\sin \left ( a+b \right )= \frac{5}{13}\times \frac{4}{5}+\frac{12}{13}\times \frac{3}{5}$
$= \frac{20}{65}+\frac{36}{65}$
$\sin \left ( a+b \right )= \frac{56}{65}$
$a+b =\sin^{-1} \frac{56}{65}$
Putting the value of $a,b$
$\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}= \sin^{-1}\frac{56}{65}$
LHS=RHS
Hence proved.