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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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If $\tan \theta=\large\frac{\sin \alpha-\cos\alpha}{\sin \alpha+\cos\alpha}$ then show that $\sin\alpha+\cos\alpha=\sqrt 2\cos\theta$

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Toolbox:
  • $\tan A-\tan B=\large\frac{\tan A-\tan B}{1+\tan A\tan B}$
  • $\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
Hence proved.
answered Apr 18, 2014 by sreemathi.v
 

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