Question

If $co{s}^{-1}\frac{p}{a}+co{s}^{-1}\frac{q}{b}=\alpha$, then $\frac{p^2}{a^2}-\frac{2pq}{ab}cos\alpha +\frac{q^2}{b^2}$ is equal to

• A. $si{n}^2\alpha$
• B. $co{s}^2\alpha$
• C. $ta{n}^2\alpha$
• D. $co{t}^2\alpha$

Answer 1

The correct option is A $si{n}^2\alpha$

$co{s}^{-1}\frac{p}{a}+co{s}^{-1}\frac{q}{b}=\alpha \Rightarrow \alpha =co{s}^{-1}(\frac{p}{a}\times \frac{q}{b}-\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}})\Rightarrow cos\alpha =\frac{p}{a}\times \frac{q}{b}-\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}}\Rightarrow {(\frac{pq}{ab}-cos\alpha )}^2=1-\frac{p^2}{a^2}-\frac{q^2}{b^2}+\frac{p^2}{a^2}\times \frac{q^2}{b^2}\Rightarrow \frac{p^2}{a^2}-\frac{2pq}{ab}cos\alpha +\frac{q^2}{b^2}=1-co{s}^2\alpha \Rightarrow \frac{p^2}{a^2}-\frac{2pq}{ab}cos\alpha +\frac{q^2}{b^2}=si{n}^2\alpha$