Question

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

Answer 1

Let ${\cos }^{-1}\frac{x}{2}=\alpha$ & ${\cos }^{-1}\frac{y}{3}=\beta$

Let $\alpha +\beta =a$

$\cos a=\cos (\alpha +\beta )$

$=\cos \alpha \cos \beta -\sin \alpha \sin \beta$

$=\frac{x}{2}\times \frac{y}{3}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{9}}=\cos a$

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

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

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

Squaring , we have

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

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

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

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