If x sin-y b cos-1 and x a cos-y b sin-1- prove that x 2 a 2-y 2 b 2-2

We #160;have #160;(xasin #952; #8722;ybcos #952;)=1Squaring #160;both #160;side, #160;we #160;have:(xasin #952; #8722;ybcos #952;)2=(1)2 ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

If x sinθ-y b...

Question

$If (\frac{x}{a} sinθ - \frac{y}{b} cosθ) = 1 and (\frac{x}{a} cosθ + \frac{y}{b} sinθ) = 1,$ prove that $(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = 2 .$

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Solution

$\begin{array}{l} We have (\frac{x}{a} \sin θ - \frac{y}{b} \cos θ) = 1 \\ Squaring both side, we have: \\ {(\frac{x}{a} \sin θ - \frac{y}{b} \cos θ)}^{2} = {(1)}^{2} \\ \Rightarrow (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ - 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 1 ... (i) \\ Again, (\frac{x}{a} \cos θ + \frac{y}{b} \sin θ) = 1 \\ Squaring both side, we get: \\ {(\frac{x}{a} \cos θ + \frac{y}{b} \sin θ)}^{2} = {(1)}^{2} \\ \Rightarrow (\frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ + 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 1 ... (ii) \\ Now, adding (i) and (ii), we get : \\ (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ - 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) + (\frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ + 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ + \frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ = 2 \\ \Rightarrow (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{x^{2}}{a^{2}} \cos^{2} θ) + (\frac{y^{2}}{b^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} (\sin^{2} θ + \cos^{2} θ) + \frac{y^{2}}{b^{2}} (\cos^{2} θ + \sin^{2} θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2 [∵ \sin^{2} θ + \cos^{2} θ = 1] \\ ∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2 \end{array}$

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Similar questions

Q. Prove the following trigonometric identities.

If

\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1 and \frac{x}{a} \sin θ - \frac{y}{b} \cos θ = 1, prove that \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2

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\frac{x^{2}}{a^{2} + λ_{1}} + \frac{y^{2}}{b^{2} + λ_{1}} = 1 and \frac{x^{2}}{a^{2} + λ_{2}} + \frac{y^{2}}{b^{2} + λ_{2}} = 1

intersect at right angles.

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If xasinθ-ybcosθ=1 and xacosθ+ybsinθ=1, prove that x2a2+y2b2=2.

$If (\frac{x}{a} sinθ - \frac{y}{b} cosθ) = 1 and (\frac{x}{a} cosθ + \frac{y}{b} sinθ) = 1,$ prove that $(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = 2 .$