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Simplest form of tan-1\((\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}+\sqrt{1-cosx}})\),π < x < \(\frac{3\pi}{2}\)is :Simplest form of tan-1(√(1+cosx)+√(1-cosx)/√(1+cosx)-√(1-cosx)),π < x < 3π/2 is :(a) \(\frac{\pi}{4}\) - \(\frac{x}{2}\)(b) \(\frac{3\pi}{2}\) - \(\frac{x}{2}\)(c) - \(\frac{x}{2}\)(d) π - \(\frac{x}{2}\) |
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Answer» Option : (a) \(tan^{-1}\left(\frac{\sqrt{1+cos\,x}+\sqrt{1-cos\,x}}{\sqrt{1+cos\,x}-\sqrt{1-cos\,x}}\right),\) \(=tan^{-1}\left(\cfrac{-\sqrt2cos\frac{x}{2}+\sqrt2sin\frac{x}{2}}{-\sqrt2cos\frac{x}{2}-\sqrt2sin\frac{x}{2}}\right),\) \(\pi<x<\frac{3\pi}{2}\) \(=tan^{-1}\left(\cfrac{cos\frac{x}{2}-sin\frac{x}{2}}{cos\frac{x}{2}+sin\frac{x}{2}}\right)\) \(=tan^{-1}\left(\cfrac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}\right)\) \(=tan^{-1}\left(\cfrac{tan\frac{\pi}{4}-tan\frac{x}{2}}{1+tan\frac{\pi}{4}tan\frac{x}{2}}\right)\) \(=tan^{-1}\left(tan(\frac{\pi}{4}-\frac{x}{2})\right)\) \(=\frac{\pi}{4}-\frac{x}{2}\) |
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