If θ is eleminated from the equation x=a cos(θ−α) and y=b cos(θ−β) then prove that (x^2 /a^2 ) + (y^2 /b^2 ) −((2xy)/(ab)) cos(α−β) = sin^2 (α−β)


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Question Number 75484 by vishalbhardwaj last updated on 11/Dec/19

$If θ is eleminated from the$ $equation x = a c o s (θ - α) and$ $y = b c o s (θ - β) then prove that$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{2 x y}{a b} c o s (α - β)$ $= {s i n}^{2} (α - β)$
	Answered by MJS last updated on 12/Dec/19

	$x = a \cos (θ - α) \Rightarrow θ = α - \arccos \frac{x}{a}$ $y = b \cos (α - β - \arccos \frac{x}{a})$ $[\cos (u + v) = \cos u \cos v - \sin u \sin v]$ $[\sin (u + v) = \cos u \sin v - \sin u \cos v]$ $y = \frac{b}{a} x \cos (α - β) \pm \frac{b}{a} \sqrt{a^{2} - x^{2}} \sin (α - β)$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{2 x y}{a b} \cos (α - β) = \sin^{2} (α - β)$ $\Rightarrow y = \frac{b}{a} x \cos (α - β) \pm \frac{b}{a} \sqrt{a^{2} - x^{2}} \sin (α - β)$ $same as above \Rightarrow proved$
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