#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.13 Question 1  Maths Textbook Solution.

Given:

$\cos^{-1}\frac{x}{2}+\cos^{-1}\frac{y}{3}= \alpha$

To prove:

$9x^{2}-12xy\cos \alpha +4y^{2}= 36\sin ^{2}\alpha$

Hint:

We are applying the formula of

$\cos^{-1}x+\cos^{-1}y= \cos^{-1}\left [ xy-\left ( \sqrt{1-x^{2}} \right )\left ( \sqrt{1-y^{2}} \right )\right ]$

Solution:

We have

$\cos^{-1}\frac{x}{2}+\cos^{-1}\frac{y}{3}= \cos^{-1}\left [ \frac{x}{2}\times \frac{y}{3}-\left ( \sqrt{1-\left ( \frac{x}{2} \right )^{2}} \right )\left ( \sqrt{1-\left ( \frac{y}{3} \right )^{2}} \right )\right ]$

$= \cos^{-1}\left [ \frac{xy}{6}-\frac{\sqrt{4-x^{2}}}{2}\times \frac{\sqrt{9-y^{2}}}{3} \right ]$

$= \cos^{-1}\left [ \frac{xy-\sqrt{4-x^{2}}\times \sqrt{9-y^{2}}}{6} \right ]= \alpha$ (Let)
$\Rightarrow xy-\sqrt{4-x^{2}} \times \sqrt{9-y^{2}} = 6\cos \alpha$
$\Rightarrow \left ( xy-6\cos \alpha \right )= \sqrt{4-x^{2}}\times \sqrt{9-y^{2}}$
On squaring both sides, we get

$\! \! \! \! \! \! \! \! \Rightarrow \left ( xy-6\cos \alpha \right )^{2}= \left ( 4-x^{2} \right )\left ( 9-y ^{2}\right )\\\Rightarrow x^{2y^{2}}+36\cos ^{2}\alpha -12xy\cos \alpha = 36-9x^{2}-4y^{2}+x^{2}y^{2}$
$\! \! \! \! \! \! \! \! \! \Rightarrow 9x^{2}+4y^{2}-36+36\cos ^{2}\alpha-12xy\cos \alpha = 0\\\Rightarrow 9x^{2}+4y^{2}-12xy\cos \alpha-36\left ( 1-\cos ^{2}\alpha \right )= 0$
$\Rightarrow 9x^{2}+4y^{2}-12xy\cos \alpha -36\sin ^{2}\alpha = 0$
$\Rightarrow 9x^{2}+4y^{2}-12xy\cos \alpha = 36\sin ^{2}\alpha$
Hence proved