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Miscellaneous Exercise 2 · Q81

Q.Prove that 2sec⁡2θ−sec⁡4θ−2 cosec2θ+cosec4θ=cot⁡4θ−tan⁡4θ2\sec^2\theta − \sec^4\theta − 2\,\text{cosec}^2\theta + \text{cosec}^4\theta = \cot^4\theta − \tan^4\theta.

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Step 1. Let s=sec⁡2θ=1+tan⁡2θs=\sec^2\theta=1+\tan^2\theta, c=cosec2θ=1+cot⁡2θc=\text{cosec}^2\theta=1+\cot^2\theta.

Step 2. LHS =2s−s2−2c+c2=(c2−s2)−2(c−s)=(c−s)(c+s−2)=2s-s^2-2c+c^2 = (c^2-s^2)-2(c-s) = (c-s)(c+s-2).

Step 3. c−s=cot⁡2θ−tan⁡2θc-s = \cot^2\theta-\tan^2\theta, and c+s−2=cot⁡2θ+tan⁡2θc+s-2 = \cot^2\theta+\tan^2\theta (since c+s=2+cot⁡2θ+tan⁡2θc+s=2+\cot^2\theta+\tan^2\theta). …

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