Question

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

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

Answer 1

The Correct Option is A

Explanation:
Given,$\frac{p^2}{a^2}+k\cos \alpha +\frac{q^2}{b^2}={\sin }^2\alpha ......$(i)${\cos }^{-1}\left(\frac{p}{a}\right)+{\cos }^{-1}\left(\frac{q}{b}\right)=\alpha$As we know,${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}(xy-\sqrt{1-x^2}\cdot \sqrt{1-y^2})$${\cos }^{-1}(\frac{pq}{ab}-\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}})=\alpha$$\cos \alpha =(\frac{pq}{ab}-\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}})$$\frac{pq}{ab}-\cos \alpha =\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}}$Squaring both sides, we get$(\frac{pq}{ab}-\cos \alpha {)}^2=(\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}}{)}^2$$\frac{(pq{)}^2}{(ab{)}^2}+{\cos }^2\alpha -2\frac{pq}{ab}\cos \alpha =(1-\frac{p^2}{a^2})(1-\frac{q^2}{b^2})$$\frac{(pq{)}^2}{(ab{)}^2}+{\cos }^2\alpha -2\frac{pq}{ab}\cos \alpha =1-\frac{p^2}{a^2}-\frac{q^2}{b^2}+\frac{(pq{)}^2}{(ab{)}^2}$${\sin }^2\alpha =\frac{p^2}{a^2}+\frac{q^2}{b^2}-2\frac{pq}{ab}\cos \alpha .....$(ii)Comparing equation (i) and (ii), we get$\mathrm{k}=-\frac{2pq}{ab}$Hence, the correct option is (A).