Question

Prove that:-

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

Answer 1

L.H.S$=\frac{sin\theta -cos\theta +1}{sin\theta +cos\theta -1}$ [divide by $cos\theta$ both number and denemenotor]

$\frac{\frac{sin\theta -cos\theta +1}{cos\theta }}{\frac{sin\theta +cos\theta -1}{cos\theta }}=\frac{\frac{sin\theta }{cos\theta }-1+\frac{1}{cos\theta }}{\frac{sin\theta }{cos\theta }+1-\frac{1}{cos\theta }}$

$=\frac{tan\theta -1+sec\theta }{tan\theta +1-sec}$ $\because [tan\theta =\frac{sin\theta }{cos\theta };sec=\frac{1}{cos\theta }]$

we know that, $se{c}^2\theta -ta{n}^2\theta =1...\left(1\right)$

$=\frac{tan\theta +sec\theta -(se{c}^2\theta -ta{n}^2\theta )}{tan\theta +1-sec\theta }$

$=\frac{(tan\theta +sec\theta )-(sec\theta -tan\theta )(sec\theta +tan\theta )}{tan\theta -sec\theta +1}$

$=\frac{(sec\theta +tan\theta )(1-sec\theta +tan\theta )}{(tan\theta -sec\theta +1)}$

$=sec\theta +tan\theta$

(multiply and divide by $sec\theta -tan\theta )$

$=\frac{(sec\theta +tan\theta )(sec\theta +tan\theta )}{(sec\theta -tan\theta )}$

$=\frac{se{c}^2\theta -ta{n}^2\theta }{sec\theta -tan\theta }$

$=\frac{1}{sec\theta -tan\theta }$ using eq (1)

Therefore, LHS = RHS

Hence, proved