Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${cos}^{-1}\left(p/a\right)+{cos}^{-1}\left(q/b\right)=\alpha$, then
$\frac{p^2}{a^2}-\frac{2pq}{ab}cos\alpha +\frac{q^2}{b^2}={sin}^2\alpha$

• A. True
• B. False

Answer 1

The correct option is A True

Given

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

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

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

$⟹\cos \alpha =\frac{p}{q}\times \frac{q}{b}-\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}}$

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

Squaring on both sides

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

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

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