Question

If ${\cos }^{-1}\frac{p}{a}+{\cos }^{-1}\frac{q}{b}=\alpha$, then prove that $\frac{p^2}{a^2}-\frac{2pq}{ab}\cos \alpha +\frac{q^2}{b^2}={\sin }^2\alpha$

Answer 1

Let $\cos A=\frac{p}{a}$ and $\cos B=\frac{q}{b}$

Thus $\alpha =A+B$

$\Rightarrow \cos \alpha =\cos (A+B)$ (Taking $\cos$ on both sides)

$\Rightarrow \cos \alpha =\cos A\cos B-\sin A\sin B$

$\Rightarrow \cos \alpha =\frac{p}{a}\frac{q}{b}-\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}}$

$\Rightarrow \cos \alpha =\frac{pq}{ab}-\frac{\sqrt{(a^2-p^2)(b^2-q^2)}}{ab}$

$\Rightarrow \cos \alpha =\frac{pq-\sqrt{(a^2-p^2)(b^2-q^2)}}{ab}$

$\Rightarrow {\cos }^2\alpha ={\left(\frac{pq-\sqrt{(a^2-p^2)(b^2-q^2)}}{ab}\right)}^2$ (squaring both sides)

$\Rightarrow {\cos }^2\alpha =\frac{p^2{q}^2-2pq\sqrt{(a^2-p^2)(b^2-q^2)}+(a^2-p^2)(b^2-q^2)}{a^2{b}^2}$

$\Rightarrow {\cos }^2\alpha =\frac{a^2{b}^2-b^2{p}^2-a^2{q}^2+2p^2{q}^2-2pq\sqrt{(a^2-p^2)(b^2-q^2)}}{a^2{b}^2}$

Lets take $1-{\cos }^2\alpha =1-\frac{a^2{b}^2-b^2{p}^2-a^2{q}^2+2p^2{q}^2-2pq\sqrt{(a^2-p^2)(b^2-q^2)}}{a^2{b}^2}$

$\Rightarrow {\sin }^2\alpha =\frac{a^2{b}^2-a^2{b}^2+b^2{p}^2+a^2{q}^2-2p^2{q}^2+2pq\sqrt{(a^2-p^2)(b^2-q^2)}}{a^2{b}^2}$

$\Rightarrow {\sin }^2\alpha =\frac{b^2{p}^2+a^2{q}^2}{a^2{b}^2}-\frac{2p^2{q}^2-2pq\sqrt{(a^2-p^2)(b^2-q^2)}}{a^2{b}^2}$

$\Rightarrow {\sin }^2\alpha =\frac{p^2}{a^2}+\frac{q^2}{b^2}-\frac{2pq}{ab}\times \frac{pq-\sqrt{(a^2-p^2)(b^2-q^2)}}{ab}$

$\Rightarrow {\sin }^2\alpha =\frac{p^2}{a^2}+\frac{q^2}{b^2}-\frac{2pq}{ab}\cos \alpha$ (substituting value of $\cos \alpha$ found above)

Hence, proved.