Question

If $co{s}^{-1}\left(\frac{p}{a}\right)+co{s}^{-1}\left(\frac{q}{b}\right)=$then $\frac{p^2}{a^2}+kcos\alpha +\frac{q^2}{b}=si{n}^2\alpha$, where k is equal to

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is B

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