#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 1 Subquestion (ii) Maths Textbook Solution.

$\pi$
Hint:
To solve this type of question, we must convert $\sin$ and $\cos$ in $\tan$ using hypotenuse theorem.
Given:
$\sin^{-1}\left ( \frac{12}{13} \right )+\cos^{-1}\left ( \frac{4}{5} \right )+\tan^{-1}\left ( \frac{63}{16} \right )=\pi$
Explanation:
For proving this type of question,
Let’s assume,
$\! \! \! \! \! \! \! \! \sin^{-1}\left ( \frac{12}{13} \right )= \alpha \\\cos^{-1}\left ( \frac{4}{5} \right )= \beta \\\tan^{-1}\left ( \frac{63}{16} \right )=\gamma$
So, we have to prove that
$\alpha+ \beta+ \gamma = \pi$
First of all
$\sin\alpha =\frac{12}{13}$

Using the formula of hypotenuse , third side will be 5.
$\tan\alpha =\frac{12}{5} \cdot \cdot \cdot (i)$
Also we have,
$\cos\beta =\frac{4}{5}$

Using the formula of hypotenuse, third side will be 3
$\tan\beta =\frac{3}{4}$                                                                                                     …(ii)
Also we have,
$\! \! \! \! \! \! \! \! \tan \gamma =\frac{63}{16} \cdot \cdot \cdot \cdot (iii)\\ \alpha +\beta +\gamma =\pi \\ \alpha +\beta =\pi -\gamma \\$
Let’s take tan on both sides,
$\tan\alpha +\beta =\tan\pi -\gamma$
$\frac{\tan \alpha +\tan \beta}{1-\ tan\alpha \tan \beta } =\frac{\tan\pi +\tan\gamma }{ 1- \tan\pi \tan\gamma }$
Let’s put value, from (i), (ii) and (iii)
\begin{aligned} &\Rightarrow \frac{\frac{12}{5}+\frac{3}{4}}{1-\left(\frac{12}{5} \times \frac{3}{4}\right)}=\frac{\tan \pi-\tan \gamma}{1-\tan \gamma \tan \pi} \\ &\Rightarrow \frac{\frac{12}{5}+\frac{3}{4}}{1-\left(\frac{12}{5} \times \frac{3}{4}\right)}=\frac{-63}{16} \\ &\Rightarrow \frac{48+15}{-16}=\frac{-63}{16} \\ &\Rightarrow \frac{-63}{16}=\frac{-63}{16} \end{aligned}
∴ L.H.S =R.H.S
Hence, the prove
Note: This type of problem can be easier by $\alpha+ \beta+ \gamma = \pi$