If x/a - y/b\;θ = 1 and x/a\;θ + y/b = 1, then the value of x²/a² + y²/b² is [2016 CDS-I]

Correct Answer is : 2cos⁡2θ2²θ2cos2θ

Correct Answer is : 2cos⁡2θ2²θ2cos2θ

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UPSC CDS Math Trigonometric Ratios and Trigonometric Identities Questions

Question : 57 of 107

Marks: +1, -0

\frac{x}{a} - \frac{y}{b}\;\tan\theta = 1

\frac{x}{a}\;\tan\theta + \frac{y}{b} = 1,

\frac{x^2}{a^2} + \frac{y^2}{b^2}

Solution:

\frac{x}{a} - \frac{y}{b}\tan\theta = 1

........(1)

\frac{x}{a}\tan\theta + \frac{y}{b} = 1

........(2)

Squaring and adding (1) and (2), we have:

\left( \frac{x}{a} - \frac{y}{b}\tan\theta \right)^2 + \left( \frac{x}{a}\tan\theta + \frac{y}{b} \right)^2

= 1 + 1 = 2

\Rightarrow\;\; \frac{x^{2}}{a^{2}} + \;\; \frac{y^{2}}{b^{2}}\tan^{2}\theta - 2\;\; \frac{xy}{ab}\tan\theta

+\;\; \frac{x^{2}}{a^{2}}\tan^{2}\theta + \;\; \frac{y^{2}}{b^{2}} + 2\;\; \frac{xy}{ab}\tan\theta = 2

\Rightarrow \left( \;\; \frac{x^{2}}{a^{2}} + \;\; \frac{y^{2}}{b^{2}} \right) \left( 1 + \tan^{2}\theta \right) = 2

\Rightarrow\;\; \frac{x^{2}}{a^{2}} + \;\; \frac{y^{2}}{b^{2}} = \;\; \frac{2}{\sec^{2}\theta} = 2\cos^{2}\theta

Alternative method:

On solving (1) and (2), we get

\;\; \frac{x}{a} = \;\; \frac{1+\tan\theta}{\sec^{2}\theta}, \;\; \frac{y}{b} = \;\; \frac{1-\tan\theta}{\sec^{2}\theta}

\therefore\;\; \frac{x^{2}}{a^{2}} + \;\; \frac{y^{2}}{b^{2}} = \;\; \frac{(1+\tan\theta)^{2} + (1-\tan\theta)^{2}}{\sec^{4}\theta}

= \;\; \frac{2\sec^{2}\theta}{\sec^{4}\theta} = 2\cos^{2}\theta

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