Question

If ${\cos }^{-1}\left(\frac{p}{a}\right)+{\cos }^{-1}\left(\frac{p}{b}\right)=\alpha$, then $\frac{p^2}{a^2}+k\cos \alpha +\frac{p^2}{b^2}={\sin }^2\alpha$, where $k$ is equal to

• A. $\frac{2pq}{ab}$
• B. $-\frac{2pq}{ab}$
• C. $-\frac{pq}{ab}$
• D. $\frac{pq}{ab}$

Answer 1

Answer: (B)

$-\frac{2pq}{ab}$

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

$\Rightarrow {\cos }^{-1}[\frac{p}{a}.\frac{q}{b}-\sqrt{(1-\frac{p^2}{a^2})}\sqrt{(1-\frac{q^2}{b^2})}]=\alpha$

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

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

$\Rightarrow \frac{p^2}{a^2}-\frac{2pq}{ab}\cos \alpha +\frac{q^2}{b^2}=1-{\cos }^2\alpha ={\sin }^2\alpha$
But $\frac{p^2}{a^2}+k\cos \alpha +\frac{q^2}{b^2}={\sin }^2\alpha$

$\Rightarrow k=-\frac{2pq}{ab}$