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Miscellaneous Exercise 2 · Q88

Q.Prove that (sin⁡θ+cosec θ)2+(cos⁡θ+sec⁡θ)2=tan⁡2θ+cot⁡2θ+7(\sin\theta + \text{cosec}\,\theta)^2 + (\cos\theta + \sec\theta)^2 = \tan^2\theta + \cot^2\theta + 7.

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Step 1. Expand: (sin⁡θ+cosec θ)2+(cos⁡θ+sec⁡θ)2=sin⁡2θ+2+cosec2θ+cos⁡2θ+2+sec⁡2θ(\sin\theta+\text{cosec}\,\theta)^2+(\cos\theta+\sec\theta)^2 = \sin^2\theta+2+\text{cosec}^2\theta+\cos^2\theta+2+\sec^2\theta (since sin⁡θ⋅cosec θ=1\sin\theta\cdot\text{cosec}\,\theta=1 and cos⁡θ⋅sec⁡θ=1\cos\theta\cdot\sec\theta=1, each cross term is 2). …

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