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Question:

If $\cos ^{-1} \frac{x}{2}+\cos ^{-1} \frac{y}{3}=\alpha$, then prove that $9 x^{2}-12 x y \cos \alpha+4 y^{2}=36 \sin ^{2} \alpha$

Solution:

We know

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

Now,

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

$\Rightarrow \cos ^{-1}\left[\frac{x}{2} \frac{y}{3}-\sqrt{1-\frac{x^{2}}{4}} \sqrt{1-\frac{y^{2}}{3}}\right]=\alpha$

$\Rightarrow \frac{x}{2} \frac{y}{3}-\sqrt{1-\frac{x^{2}}{4}} \sqrt{1-\frac{y^{2}}{3}}=\cos \alpha$

$\Rightarrow x y-\sqrt{4-x^{2}} \sqrt{9-y^{2}}=6 \cos \alpha$

$\Rightarrow \sqrt{4-x^{2}} \sqrt{9-y^{2}}=x y-6 \cos \alpha$

$\Rightarrow\left(4-x^{2}\right)\left(9-y^{2}\right)=x^{2} y^{2}+36 \cos ^{2} \alpha-12 x y \cos \alpha \quad$ [Squaring both sides]

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

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

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

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

 

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