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 $