Question

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

Answer 1

Consider the given equation.

${\cos }^{-1}\left(\frac{p}{a}\right)+{\cos }^{-1}\left(\frac{q}{b}\right)=\alpha$

We know that

${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}\{xy-\sqrt{1-x^2}\sqrt{1-y^2}\}$

Thus,

${\cos }^{-1}\{\frac{p}{a}\times \frac{q}{b}-\sqrt{1-{\left(\frac{p}{a}\right)}^2}\sqrt{1-{\left(\frac{q}{b}\right)}^2}\}=\alpha$

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

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

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

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

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

On squaring both sides, we get

$p^2{q}^2+a^2{b}^2{\cos }^2\alpha -2abpq\cos \alpha =(a^2-p^2)(b^2-q^2)$

$p^2{q}^2+a^2{b}^2{\cos }^2\alpha -2abpq\cos \alpha =a^2{b}^2-a^2{q}^2-b^2{p}^2+p^2{q}^2$

$a^2{b}^2{\cos }^2\alpha -2abpq\cos \alpha =a^2{b}^2-a^2{q}^2-b^2{p}^2$

$a^2{q}^2+b^2{p}^2-2abpq\cos \alpha =a^2{b}^2(1-{\cos }^2\alpha )$

$\frac{a^2{q}^2+b^2{p}^2-2abpq\cos \alpha }{a^2{b}^2}={\sin }^2\alpha$

$\frac{q^2}{b^2}+\frac{p^2}{a^2}-\frac{2pq}{ab}\cos \alpha ={\sin }^2\alpha$

$\frac{p^2}{a^2}-\frac{2pq}{ab}\cos \alpha +\frac{q^2}{b^2}={\sin }^2\alpha$

Hence, proved.