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Exercise 3.3 · Q64

Q.Prove the following: tan⁡−1(cos⁡θ+sin⁡θcos⁡θ−sin⁡θ)=π4+θ\tan^{-1}\left(\dfrac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\right) = \dfrac{\pi}{4} + \theta, if θ∈(−π4,π4)\theta \in \left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Dividing numerator and denominator by cos⁡θ\cos\theta: cos⁡θ+sin⁡θcos⁡θ−sin⁡θ=1+tan⁡θ1−tan⁡θ=tan⁡π4+tan⁡θ1−tan⁡π4tan⁡θ=tan⁡(π4+θ)\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\dfrac{1+\tan\theta}{1-\tan\theta}=\dfrac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta}=\tan\left(\dfrac{\pi}{4}+\theta\right). For θ∈(−π4,π4)\theta\in\left(-\dfrac{\pi}{4},\dfrac{\pi}{4}\right), π4+θ∈(0,π2)⊂(−π2,π2)\dfrac{\pi}{4}+\theta\in\left(0,\dfrac{\pi}{2}\right)\subset\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right), so taking tan⁡−1\tan^{-1} of both sides directly recovers the angle: $\tan^{ …

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