Question

If \[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\] , then show that \[\sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\].

Answer 1

\[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\]
Dividing numerator and denominator on the RHS by \[\cos\alpha\], we get 

\[\tan\theta = \frac{\frac{\sin\alpha}{\cos\alpha} - 1}{\frac{\sin\alpha}{\cos\alpha} + 1}\]

\[ \Rightarrow \tan\theta = \frac{\tan\alpha - \tan\frac{\pi}{4}}{1 + \tan\alpha \tan\frac{\pi}{4}}\]

\[ \Rightarrow \tan\theta = \tan\left( \alpha - \frac{\pi}{4} \right)\]

\[ \Rightarrow \theta = \alpha - \frac{\pi}{4}\]

\[\text{ Or }\alpha = \frac{\pi}{4} + \theta\]
Now,
\[\sin\alpha + \cos\alpha\]
\[ = \sin\left( \frac{\pi}{4} + \theta \right) + \cos\left( \frac{\pi}{4} + \theta \right)\]
\[ = \sin\frac{\pi}{4}\cos\theta + \cos\frac{\pi}{4}\sin\theta + \cos\frac{\pi}{4}\cos\theta - \sin\frac{\pi}{4}\sin\theta\]
\[ = \frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta - \frac{1}{\sqrt{2}}\sin\theta\]
\[ = \frac{2}{\sqrt{2}}\cos\theta\]
\[ = \sqrt{2}\cos\theta\]
\[\therefore \sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\]