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# If $\cos^{-1}\Large\frac{x}{2}+\normalsize\cos^{-1}\large\frac{y}{3}=\theta ,\;\normalsize prove\;that\;9x^2-12xy\cos\theta +4y^2=36\sin^2\theta$

Toolbox:
• $sin(cos^{-1}x)=\sqrt{1-x^2}=cos(sin^{-1}x)$
• $cos (A+B)=cosa\: cosB-sina\: sin B$
• $cos^{-1}cosx=x=cos\: cos^{-1}x$
Taking cos on both sides $cos\theta = cos \bigg[ cos^{-1}\large\frac{x}{2}+cos^{-1}\large\frac{y}{3} \bigg]$
$\Rightarrow cos\theta = \large\frac{x}{2}.\large\frac{y}{3}-\sqrt{1-\large\frac{x^2}{4}} \sqrt{1-\large\frac{y^2}{9}}$
$cos\theta-\large\frac{xy}{6}=-\large\frac{\sqrt{4-x^2}.{9-y^2}}{6}$
Squaring both the sides and simplifying which you get the proof.

edited Mar 19, 2013