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Prove that $\large\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}=$$\tan\large\frac{\theta}{2}$

1 Answer

Toolbox:
  • $\sin^2\large\frac{\theta}{2}=\frac{1}{2}$$(1-\cos 2\theta)$
  • $\cos^2\large\frac{\theta}{2}=\frac{1}{2}$$(1+\cos 2\theta)$
  • $\sin^2\theta=2\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
LHS
$\large\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}$
$\large\frac{1-\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}=\frac{2\sin ^2\Large\frac{\theta}{2}+2\sin \Large\frac{\theta}{2}.\cos\large\frac{\theta}{2}}{2\cos ^2\Large\frac{\theta}{2}+2\sin \Large\frac{\theta}{2}.\cos\large\frac{\theta}{2}}$
$\Rightarrow \large\frac{2\sin \large\frac{\theta}{2}(\sin \Large\frac{\theta}{2}+\cos\Large\frac{\theta}{2})}{2\sin \large\frac{\theta}{2}(\sin \Large\frac{\theta}{2}+\cos\Large\frac{\theta}{2})}$
$\Rightarrow \tan\large\frac{\theta}{2}$
Hence proved
answered May 22, 2014 by sreemathi.v
 

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