If $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$, then show that $\sin \alpha+\cos \alpha=\sqrt{2} \cos \theta$. [NCERT EXEMPLER]
$\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$
Dividing numerator and denominator on the RHS by $\cos \alpha$, we get
$\tan \theta=\frac{\frac{\sin \alpha}{\cos \alpha}-1}{\frac{\sin \alpha}{\cos \alpha}+1}$
$\Rightarrow \tan \theta=\frac{\tan \alpha-\tan \frac{\pi}{4}}{1+\tan \alpha \tan \frac{\pi}{4}}$
$\Rightarrow \tan \theta=\tan \left(\alpha-\frac{\pi}{4}\right)$
$\Rightarrow \theta=\alpha-\frac{\pi}{4}$
Or $\alpha=\frac{\pi}{4}+\theta$
Now,
$\sin \alpha+\cos \alpha$
$=\sin \left(\frac{\pi}{4}+\theta\right)+\cos \left(\frac{\pi}{4}+\theta\right)$
$=\sin \frac{\pi}{4} \cos \theta+\cos \frac{\pi}{4} \sin \theta+\cos \frac{\pi}{4} \cos \theta-\sin \frac{\pi}{4} \sin \theta$
$=\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta$
$=\frac{2}{\sqrt{2}} \cos \theta$
$=\sqrt{2} \cos \theta$
$\therefore \sin \alpha+\cos \alpha=\sqrt{2} \cos \theta$
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