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Miscellaneous Exercise 2 · Q76

Q.If sin⁡θ=x2−y2x2+y2\sin\theta = \dfrac{x^2−y^2}{x^2+y^2} then find the values of cos⁡θ\cos\theta, tan⁡θ\tan\theta in terms of xx and yy.

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Step 1. sin⁡θ=x2−y2x2+y2\sin\theta=\dfrac{x^2-y^2}{x^2+y^2}, so 1−sin⁡2θ=(x2+y2)2−(x2−y2)2(x2+y2)21-\sin^2\theta = \dfrac{(x^2+y^2)^2-(x^2-y^2)^2}{(x^2+y^2)^2}.

Step 2. The numerator is a difference of squares: [(x2+y2)−(x2−y2)][(x2+y2)+(x2−y2)]=(2y2)(2x2)=4x2y2\left[(x^2+y^2)-(x^2-y^2)\right]\left[(x^2+y^2)+(x^2-y^2)\right] = (2y^2)(2x^2)=4x^2y^2.

Step 3. So cos⁡2θ=4x2y2(x2+y2)2⇒cos⁡θ=±2xyx2+y2\cos^2\theta = \dfrac{4x^2y^2}{(x^2+y^2)^2} \Rightarrow \cos\theta = \pm\dfrac{2xy}{x^2+y^2}. …

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