Question

$\frac{sinq - cosq + 1}{sinq + cosq - 1} = \frac{1}{secq - tanq}$
​

Answer 1

Since we need secQ and tanQ in RHS we divide cosQ in whole RHS

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

Solution-

LHS

Divide cosQ in both numerator and denominator

$\rm \frac{ \sin \theta - \cos \theta + 1}{ \sin \theta + \cos \theta - 1} \implies \frac{ \frac{ \sin \theta}{ \cos \theta} - \frac{ \cos \theta}{ \cos \theta} + \frac{1}{ \cos \theta} }{ \frac{ \sin \theta}{ \cos \theta} + \frac{ \cos \theta}{ \cos \theta} - \frac{1}{ \cos \theta} } \\ \rm \implies \frac{ \tan \theta - 1 + \sec \theta}{ \tan \theta + 1 - \sec \theta}$

Now we multiply tanQ-secQ in both numerator and denominator

$\frac{(( \tan \theta + \sec \theta) - 1)( \tan \theta - \sec \theta)}{(( \tan \theta - \sec \theta) + 1)( \tan \theta - \sec \theta)} \\ \implies \frac{( \tan \theta + \sec \theta)( \tan \theta - \sec \theta) - ( \tan \theta - \sec \theta) }{(( \tan \theta - \sec \theta) + 1)( \tan \theta - \sec \theta)} \\ \rm \implies \frac{ {tan}^{2} \theta - {sec}^{2} \theta - tan \theta + sec \theta }{( \tan \theta - \sec \theta) + 1)( \tan \theta - \sec \theta)} \\ \rm \implies \frac{ - (1 + tan \theta - sec \theta)}{( \tan \theta - \sec \theta) + 1)( \tan \theta - \sec \theta)} \\ \implies \rm \frac{ - 1}{tan \theta - sec \theta} = \frac{1}{sec \theta - tan \theta}$

$\rm LHS = RHS \\ \rm hence \: proved$