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Relation between circular and hyperbolic function : Prove that $sec \;h^{-1}(sin\theta) = log(cot(\theta/2))$
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Let, $sec \; h^{-1}(sin\theta) = z sec \; hz \; = \; sin \theta \\ \therefore \frac{1}{cosh \; z} = sin\theta \; \; \; \therefore \frac{1}{\frac{e^z + e^{-z}}{2}} = sin\theta \\ \; \\ \therefore \frac{2}{e^z + \frac{1}{e^z}} = sin \theta \; \; \; \therefore \frac{2e^z}{e^{2z}+1} = sin\theta \\ \; \\ \therefore e^{2z} sin\theta - 2e^z + sin\theta = 0 \\ \; \\ \therefore e^z = \tfrac{ -(-2) \; \pm \; \sqrt{ (-2)^2 \; - \; 4sin\theta \; \cdot \; sin\theta } }{ 2sin\theta } \; \; = \; \; \tfrac{ 2 \; \pm \; \sqrt{ 4 \; - \; 4sin^2\theta } }{2sin\theta} \\ \; \\ \therefore e^z = \tfrac{ 2 \; \pm \; 2( \sqrt{ 1 \; - \; sin^2\theta) } }{2sin\theta} \; \; = \; \; \tfrac{ (1 \; \pm \; \sqrt{cos^2\theta) } }{sin\theta} \; \; = \; \; \frac{1 \; \pm \; cos\theta}{sin\theta} \\ \; \\ We \; \; take \; \; e^z = \frac{1+cos\theta}{sin\theta} \; = \; \tfrac{ 2cos^2 \frac{\theta}{2}}{2 sin\frac{\theta}{2} \; cos \frac{\theta}{2}} \\ \; \\ \therefore e^z = cot\frac{\theta}{2} \\ \; \\ By \; \; taking \; \; log \; \; on \; \; both \; \; sides, \\ \; \\ \therefore z = log(cot\frac{\theta}{2}) \\ \; \\ \therefore sec \; h^{-1}(sin\theta) = log(cot(\theta/2)) \; \; \; \ldots \ldots Hence \; \; Proved $

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