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Miscellaneous Exercise 2 · Q87

Q.Prove that tan⁡2θ−sin⁡2θ=sin⁡4θsec⁡2θ\tan^2\theta − \sin^2\theta = \sin^4\theta\sec^2\theta.

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Step 1. tan⁡2θ−sin⁡2θ=sin⁡2θcos⁡2θ−sin⁡2θ=sin⁡2θ(1cos⁡2θ−1)=sin⁡2θ⋅1−cos⁡2θcos⁡2θ\tan^2\theta-\sin^2\theta = \dfrac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta = \sin^2\theta\left(\dfrac1{\cos^2\theta}-1\right) = \sin^2\theta\cdot\dfrac{1-\cos^2\theta}{\cos^2\theta}. …

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