Question

If ${\cos }^{-1}\left(\frac{x}{a}\right)+{\cos }^{-1}\left(\frac{y}{b}\right)=\alpha ,$ then the value of $\frac{x^2}{a^2}-\frac{2xy}{ab}\cos \alpha +\frac{y^2}{b^2}$ is :

• A. ${\sin }^2\alpha$
• B. ${\cos }^2\alpha$
• C. ${\tan }^2\alpha$
• D. ${\cot }^2\alpha$

Answer 1

Answer: (A)

Note : ${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}$
$\left[xy-\sqrt{1-x^2}\sqrt{1-y^2}\right]$
Let ${\cos }^{-1}\left(\frac{x}{a}\right)+{\cos }^{-1}\left(\frac{y}{b}\right)=\alpha$
Now, replace x by $\frac{x}{a}$ and y by $\frac{y}{b}$ in the given note, we get
${\cos }^{-1}\left[\frac{x}{a},\frac{y}{b}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}\right]$
$={\cos }^{-1}\left(\frac{x}{a}\right)+{\cos }^{-1}\left(\frac{y}{b}\right)=\alpha$
$\Rightarrow \frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}=\cos \alpha$
$\Rightarrow \frac{xy}{ab}-\sqrt{\frac{a^2-x^2}{a^2}}\sqrt{\frac{b^2-y^2}{b^2}}=\cos \alpha$
$\Rightarrow \frac{xy}{ab}-\frac{1}{ab}\sqrt{a^2-x^2}\sqrt{b^2-y^2}=\cos \alpha$
$\Rightarrow \frac{xy}{ab}-\cos \alpha =\frac{1}{ab}\sqrt{a^2-x^2}\sqrt{b^2-y^2}$
Now, square both side, we get
${\left(\frac{xy}{ab}-\cos \alpha \right)}^2=\frac{1}{a^2{b}^2}\left(a^2-x^2\right)\left(b^2-y^2\right)$
$\Rightarrow \frac{x^2{y}^2}{a^2{b}^2}+{\cos }^2\alpha -\frac{2xy}{ab}\cos \alpha$
$=1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2{y}^2}{a^2{b}^2}$
$\Rightarrow \frac{x^2}{a^2}-\frac{2xy}{ab}\cos \alpha +\frac{y^2}{b^2}$
$=1-{\cos }^2\alpha ={\sin }^2\alpha$