Question

Prove that: $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\sec \theta +\tan \theta$

Answer 1

$LHS=\left(\frac{sin\theta -cos\theta +1}{sin\theta +cos\theta -1}\right)\times \left(\frac{sin\theta +cos\theta +1}{sin\theta +cos\theta +1}\right)$

$=\left(\frac{(si{n}^2\theta +1{)}^2-co{s}^2\theta }{(sin\theta +cos\theta {)}^2-1}\right)$

$=\left(\frac{si{n}^2\theta +2sin\theta +1-co{s}^2\theta }{si{n}^2\theta +co{s}^2\theta +2sin\theta \cos \theta -1}\right)$

$=\left(\frac{si{n}^2\theta +2sin\theta +1-(1-si{n}^2\theta )}{2sin\theta \cos \theta }\right)$

$=\left(\frac{2si{n}^2\theta +2sin\theta }{2sin\theta \cos \theta }\right)$

$=\left(\frac{sin\theta +1}{cos\theta }\right)$

$=tan\theta +sec\theta =RHS$