#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 13 Maths Textbook Solution.

Answer: $18-18 \cos \theta$

Hint: Take cos function to the RHS, so the variables get free.

Given: $\cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}, \text { then } 4 x^{2}-12 x y \cos \frac{\theta}{2}+9 y^{2}=?$

Solution:

\begin{aligned} &\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) \\ &\cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2} \\ &\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)=\frac{\theta}{2} \\ &\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)=\cos \frac{\theta}{2} \\ &x y-6 \cos \frac{\theta}{2}=\sqrt{9-x^{2}} \sqrt{4-y^{2}} \end{aligned}

Squaring on both sides,

\begin{aligned} &x^{2} y^{2}-12 x y \cos \frac{\theta}{2}+36 \cos ^{2} \frac{\theta}{2}=\left(9-x^{2}\right)\left(4-y^{2}\right) \\ &x^{2} y^{2}-12 x y \cos \frac{\theta}{2}+36 \cos ^{2} \frac{\theta}{2}=36-9 y^{2}-4 x^{2}+x^{2} y^{2} \\ &4 x^{2}+9 y^{2}-12 x y \cos \frac{\theta}{2}=36-36 \cos ^{2} \frac{\theta}{2} \\ &=36\left\{1-\left(\frac{1+\cos \theta}{2}\right)\right\} \end{aligned}

Here, $\therefore \cos 2 x=2 \cos ^{2} x-1$

$\Rightarrow 4 x^{2}+9 y^{2}-12 x y \cos \frac{\theta}{2}=18-18 \cos \theta$