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Question: If \(\cos^{- 1}\left( \frac{x}{a} \right) + \cos^{- 1}\left( \frac{y}{b} \right) = \alpha\), then \(...

If cos⁡−1(xa)+cos⁡−1(yb)=α\cos^{- 1}\left( \frac{x}{a} \right) + \cos^{- 1}\left( \frac{y}{b} \right) = \alpha, then x2a2−2xyabcos⁡α+y2b2=\frac{x^{2}}{a^{2}} - \frac{2xy}{ab}\cos\alpha + \frac{y^{2}}{b^{2}} =

A

sin⁡2α\sin^{2}\alpha

B

cos⁡2α\cos^{2}\alpha

C

tan⁡2α\tan^{2}\alpha

D

cot⁡2α\cot^{2}\alpha

Answer

sin⁡2α\sin^{2}\alpha

Explanation

Solution

We have cos⁡−1[xa.yb−(1−x2a2)(1−y2b2)]=α\cos^{- 1}\left\lbrack \frac{x}{a}.\frac{y}{b} - \sqrt{\left( 1 - \frac{x^{2}}{a^{2}} \right)}\sqrt{\left( 1 - \frac{y^{2}}{b^{2}} \right)} \right\rbrack = \alpha

⇒xyab−(1−x2a2)(1−y2b2)=cos⁡α\Rightarrow \frac{xy}{ab} - \sqrt{\left( 1 - \frac{x^{2}}{a^{2}} \right)}\sqrt{\left( 1 - \frac{y^{2}}{b^{2}} \right)} = \cos\alpha

∴\therefore (xyab−cos⁡α)2=1−x2a2−y2b2+x2y2a2b2\left( \frac{xy}{ab} - \cos\alpha \right)^{2} = 1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{x^{2}y^{2}}{a^{2}b^{2}}

x2y2a2b2+cos⁡2α−2xyabcos⁡α=1−x2a2−y2b2+x2y2a2b2\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}}

⇒ x2a2−2xyabcos⁡α+y2b2=1−cos⁡2α=sin⁡2α\frac{x^{2}}{a^{2}} - \frac{2xy}{ab}\cos\alpha + \frac{y^{2}}{b^{2}} = 1 - \cos^{2}\alpha = \sin^{2}\alpha