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Q.
If $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$, then $\sin \alpha+\cos \alpha$ and $\sin \alpha-\cos \alpha$ must be equal to
Trigonometric Functions
Solution:
We have $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$
$=\frac{\sin \left(\alpha-\frac{\pi}{4}\right)}{\cos \left(\alpha-\frac{\pi}{4}\right)}$
$\Rightarrow \tan \theta=\tan \left(\alpha-\frac{\pi}{4}\right)$
$\Rightarrow \theta=\alpha-\frac{\pi}{4}$
$\Rightarrow \alpha=\theta+\frac{\pi}{4}$
Put value of $\alpha$ and find $\sin \alpha+\cos \alpha=\sqrt{2} \cos \theta$ and $\sin \alpha-\cos \alpha=\sqrt{2} \sin \theta$