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# If $\tan \theta=\large\frac{\sin \alpha-\cos\alpha}{\sin \alpha+\cos\alpha}$ then show that $\sin\alpha+\cos\alpha=\sqrt 2\cos\theta$

• $\tan A-\tan B=\large\frac{\tan A-\tan B}{1+\tan A\tan B}$
• $\tan \large\frac{\pi}{4}$$=1 • \sin(A+B)=\sin A\cos B+\cos A\sin B • \cos(A+B)=\cos A\cos B+\sin A\sin B Given :\tan \theta=\large\frac{\sin \alpha-\cos\alpha}{\sin \alpha+\cos\alpha} \Rightarrow \large\frac{\Large\frac{\sin \alpha-\cos \alpha}{\cos \alpha}}{\Large\frac{\sin \alpha+\cos \alpha}{\cos \alpha}} \tan\theta=\large\frac{\Large\frac{\sin \alpha}{\cos \alpha}-1}{\Large\frac{\sin \alpha}{\cos \alpha}+1} \tan\theta=\large\frac{\tan\alpha+\tan\large\frac{\pi}{4}}{\tan\alpha\tan\large\frac{\pi}{4}+1} \tan\theta=\tan(\alpha-\large\frac{\pi}{4})$$\Rightarrow \theta=\alpha-\large\frac{\pi}{4}$
$\alpha=\large\frac{\pi}{4}$$+\theta L.H.S \sin \alpha+\cos \alpha \sin(\large\frac{\pi}{4}$$+\theta)+\cos(\large\frac{\pi}{4}$$+\theta) \Rightarrow \sin \large\frac{\pi}{4}$$.\cos\theta\cos \large\frac{\pi}{4}$$.\sin\theta+\cos \large\frac{\pi}{4}$$.\cos\theta-\sin \large\frac{\pi}{4}$$.\sin\theta \Rightarrow \cos\theta(\sin\large\frac{\pi}{4}$$+\cos\large\frac{\pi}{4})$$+\sin \theta(\cos\large\frac{\pi}{4}$$-\sin \large\frac{\pi}{4})$
$\Rightarrow \cos\theta(\large\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2})+$$\sin \theta(\large\frac{1}{\sqrt 2}-\large\frac{1}{\sqrt 2})$
$\Rightarrow \cos\theta(\large\frac{2}{\sqrt 2})$
$\Rightarrow \sqrt 2\cos\theta$=R.H.S