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Prove that $\large\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$

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Toolbox:
  • $\sec^A-\tan^2A=1$
  • $\tan A=\large\frac{\sin A}{\cos A}$
  • $\sec A=\large\frac{1}{\cos A}$
L.H.S
$\large\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$
$\large\frac{\tan A+\sec A-(\sec^2A-\tan^2A)}{\tan A-\sec A+1}$
$\large\frac{\tan A+\sec A-\big[(\sec A+\tan A)(\sec A-\tan A)\big]}{\tan A-\sec A+1}$
$\large\frac{\tan A+\sec A(1-\sec A+\tan A)}{\tan A-\sec A+1}$
$\tan A+\sec A=\large\frac{\sin A}{\cos A}+\frac{1}{\cos A}$
$\Rightarrow \large\frac{1+\sin A}{\cos A}$=R.H.S
Hence proved.
answered Apr 18, 2014 by sreemathi.v
 

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