Question

If ${\cos }^{-1}\frac{x}{3}+{\cos }^{-1}\frac{y}{2}=\frac{\theta }{2}$, then $4x^2-12xy\cos \frac{\theta }{2}+9y^2\text{is equal to:}$

• A. $36$
• B. $36-36\cos \theta$
• C. $18-18\cos \theta$
• D. $18+18\cos \theta$

Answer 1

Answer: (C)

$18-18\cos \theta$

given, ${\cos }^{-1}\frac{x}{3}+{\cos }^{-1}\frac{y}{2}=\frac{\theta }{2}$

$\Rightarrow \cos ({\cos }^{-1}\frac{x}{3}+{\cos }^{-1}\frac{y}{2})=\cos \frac{\theta }{2}$

$[\because \cos (A+B)=\cos A\cos B-\sin A\sin B]$

$\therefore \cos \left({\cos }^{-1}\frac{x}{3}\right)\cos \left({\cos }^{-1}\frac{y}{2}\right)-\sin \left({\cos }^{-1}\frac{x}{3}\right)\sin \left({\cos }^{-1}\frac{y}{2}\right)=\cos \frac{\theta }{2}$

$\Rightarrow \cos \left({\cos }^{-1}\frac{x}{3}\right)\cos \left({\cos }^{-1}\frac{y}{2}\right)-\sin \left({sin}^{-1}\sqrt{1-\frac{x^2}{9}}\right)\sin \left({\sin }^{-1}\sqrt{1-\frac{y^2}{4}}\right)=\cos \frac{\theta }{2}$

$\Rightarrow \frac{xy}{6}-\sqrt{1-\frac{x^2}{9}}\sqrt{1-\frac{y^2}{4}}=\cos \frac{\theta }{2}$

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

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

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

Squaring both sides, we get

$\Rightarrow (9-x^2)(4-y^2)=36{\cos }^2\frac{\theta }{2}+x^2{y}^2-12xy\cos \frac{\theta }{2}$

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

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

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

$\Rightarrow 9y^2+4x^2-12xy\cos \frac{\theta }{2}=36(1-{\cos }^2\frac{\theta }{2})$

$\Rightarrow 9y^2+4x^2-12xy\cos \frac{\theta }{2}=36\left({\sin }^2\frac{\theta }{2}\right)$

$\Rightarrow 9y^2+4x^2-12xy\cos \frac{\theta }{2}=18\left(2{\sin }^2\frac{\theta }{2}\right)$

$\Rightarrow 9y^2+4x^2-12xy\cos \frac{\theta }{2}=18(1-\cos \theta )$

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