$\displaystyle

\frac{tan \theta +sec \theta -1}{tan \theta - sec \theta +1}= \frac{1 + sin \theta}{cos \theta}

$

Kindly solve it covering all the steps ...:)

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- Mar 8th 2010, 07:56 PMSanjana DasProve that..
$\displaystyle

\frac{tan \theta +sec \theta -1}{tan \theta - sec \theta +1}= \frac{1 + sin \theta}{cos \theta}

$

Kindly solve it covering all the steps ...:) - Mar 8th 2010, 08:11 PMProve It
A better idea is to show us what you have done and where you are stuck.

Hint: Convert everything into a function of sines and cosines using

$\displaystyle \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

and

$\displaystyle \sec{\theta} = \frac{1}{\cos{\theta}}$. - Mar 8th 2010, 08:18 PMSanjana Das
When I solved my LHS was not coming equal to RHS n it was coming very long do u have any trick or something which can make it shorter ..

- Mar 9th 2010, 03:20 AMProve It
- Mar 9th 2010, 04:46 AMmartingoldstein
Yes. it is.