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Exercise 3.3 · Q65

Q.Prove the following: tan⁡−11−cos⁡θ1+cos⁡θ=θ2\tan^{-1}\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \dfrac{\theta}{2}, if θ∈(0,π)\theta \in (0, \pi)

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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1−cos⁡θ=2sin⁡2θ21-\cos\theta=2\sin^2\dfrac{\theta}{2} and 1+cos⁡θ=2cos⁡2θ21+\cos\theta=2\cos^2\dfrac{\theta}{2}, so 1−cos⁡θ1+cos⁡θ=tan⁡2θ2\dfrac{1-\cos\theta}{1+\cos\theta}=\tan^2\dfrac{\theta}{2}, and 1−cos⁡θ1+cos⁡θ=∣tan⁡θ2∣=tan⁡θ2\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\left|\tan\dfrac{\theta}{2}\right|=\tan\dfrac{\theta}{2} (positive, since θ∈(0,π)  ⟹  θ2∈(0,π2)\theta\in(0,\pi)\implies\dfrac{\theta}{2}\in\left(0,\dfrac{\pi}{2}\right) where tangent is positive). Since $\dfrac{\theta}{2}\in\left(0,\dfrac{\pi}{2}\right)\subset\left(-\ …

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