Prove that: \( \tan ^{-1}\left[\frac{\cos \theta+\sin \theta}{\cos \th

1.	Prove that: \( \tan ^{-1}\left[\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right]=\frac{\pi}{4}+\theta \), if \( \theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \)
Answer» tan^-1(\(\frac{cos\theta+sin\theta}{cos\theta-sin\theta}\)) = tan^-1\(\left(\cfrac{1+\frac{sin\theta}{cos\theta}}{1-\frac{sin\theta}{cos\theta}}\right)\)(By dividing both numerator and denominator by cos θ) \(=tan^{-1}(\frac{1+tan\theta}{1-tan\theta})\) \((\because\frac{sin\theta}{cos\theta}=\tan\theta)\) \(=tan^{-1}\left(\cfrac{tan\frac{\pi}4+tan\theta}{1-tan\frac{\pi}4tan\theta}\right)\)\((\because tan\frac{\pi}4=1)\) \(=tan^{-1}(tan(\frac{\pi}4+\theta))\) \((\because\frac{tanA+tanB}{1+tanAtanB}=tan(A+B))\) \(=\frac{\pi}4+\theta\) \((\because tan^{-1}(tan\theta)=\theta)\)

Prove that: \( \tan ^{-1}\left[\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right]=\frac{\pi}{4}+\theta \), if \( \theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \)

Answer»

tan^-1(\(\frac{cos\theta+sin\theta}{cos\theta-sin\theta}\)) = tan^-1\(\left(\cfrac{1+\frac{sin\theta}{cos\theta}}{1-\frac{sin\theta}{cos\theta}}\right)\)(By dividing both numerator and denominator by cos θ)

\(=tan^{-1}(\frac{1+tan\theta}{1-tan\theta})\) \((\because\frac{sin\theta}{cos\theta}=\tan\theta)\)

\(=tan^{-1}\left(\cfrac{tan\frac{\pi}4+tan\theta}{1-tan\frac{\pi}4tan\theta}\right)\)\((\because tan\frac{\pi}4=1)\)

\(=tan^{-1}(tan(\frac{\pi}4+\theta))\) \((\because\frac{tanA+tanB}{1+tanAtanB}=tan(A+B))\)

\(=\frac{\pi}4+\theta\) \((\because tan^{-1}(tan\theta)=\theta)\)

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