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Miscellaneous Exercise 2 · Q79

Q.Prove that (1+cot⁡θ+tan⁡θ)(sin⁡θ−cos⁡θ)sec⁡3θ−cosec3θ=sin⁡2θcos⁡2θ\dfrac{(1+\cot\theta+\tan\theta)(\sin\theta−\cos\theta)}{\sec^3\theta−\text{cosec}^3\theta} = \sin^2\theta\cos^2\theta.

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Step 1. 1+cot⁡θ+tan⁡θ=1+cos⁡θsin⁡θ+sin⁡θcos⁡θ=sin⁡θcos⁡θ+cos⁡2θ+sin⁡2θsin⁡θcos⁡θ=1+sin⁡θcos⁡θsin⁡θcos⁡θ1+\cot\theta+\tan\theta = 1+\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta} = \dfrac{\sin\theta\cos\theta+\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta} = \dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}.

Step 2. sec⁡3θ−cosec3θ=sin⁡3θ−cos⁡3θsin⁡3θcos⁡3θ\sec^3\theta-\text{cosec}^3\theta = \dfrac{\sin^3\theta-\cos^3\theta}{\sin^3\theta\cos^3\theta}, and by the difference-of-cubes factoring, sin⁡3θ−cos⁡3θ=(sin⁡θ−cos⁡θ)(sin⁡2θ+sin⁡θcos⁡θ+cos⁡2θ)=(sin⁡θ−cos⁡θ)(1+sin⁡θcos⁡θ)\sin^3\theta-\cos^3\theta = (\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta) = (\sin\theta-\cos\theta)(1+\sin\theta\cos\theta).

Step 3. Numerator of the whole expression: 1+sin⁡θcos⁡θsin⁡θcos⁡θ⋅(sin⁡θ−cos⁡θ)\dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}\cdot(\sin\theta-\cos\theta). …

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