Question

pls solve this solution in attachment

pls...​

Answer 1

Trigonometric Identity:

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

$\sf{Divide\:numerator\:and\:denominator\:by\:cos\theta}$

$\implies\Large\sf{\frac{ \frac{sin \theta}{cos \theta} - \frac{cos \theta}{cos \theta} + \frac{1}{cos \theta} }{\frac{sin \theta}{cos \theta} + \frac{cos \theta}{cos \theta} - \frac{1}{cos \theta}}}$

$\sf{\implies \frac{tan \theta - 1 + sec \theta}{tan \theta + 1 - sec \theta}}$

$\sf{Rationalize \:the \:denominator}$

$\sf{\implies \frac{(tan \theta - 1) + (sec \theta)}{(tan \theta + 1) - (sec \theta)} \times \frac{(tan \theta + 1) + (sec \theta)}{(tan \theta + 1) + (sec \theta)}}$

$\Large\sf{\implies \frac{ \binom{ {tan}^{2} \theta + \cancel{tan \theta} + tan \theta sec \theta \cancel{ - tan \theta} - 1 }{ \cancel{- sec \theta} + tan \theta sec \theta +\cancel{sec \theta} + {sec}^{2} \theta} }{ {tan}^{2} \theta + 1 + 2tan \theta - {sec}^{2} \theta}}$

$\sf{\implies \frac{{tan}^{2} \theta + 2 tan\theta sec \theta - 1 + {sec}^{2} }{ \cancel{{sec }^{2} } \theta + 2tan \theta \cancel{ - {sec}^{2} \theta}} }$

$\sf{\implies \frac{2 {tan}^{2} \theta + 2tan \theta sec \theta}{2tan \theta}}$

$\sf{\implies \frac{ \cancel{2tan \theta}(tan \theta + sec \theta)}{ \cancel{2tan \theta}} }$

$\sf{\implies tan\theta+sec\theta}$

$\sf{Multiply \:numerator \:and\:denominator \:by}$

$\sf{ tan\theta-sec\theta}$

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

$\sf{\implies \frac{{tan}^{2} \theta - {sec }^{2} \theta}{tan \theta - sec \theta} }$

$\sf{\implies \frac{ - 1}{tan \theta - sec \theta} }$

$\sf{\implies\frac{1}{sec \theta - tan \theta} =RHS}$

________________

@zaqwertyuioplm

Answer 2

here is your ans== see in attachment. ..

...........however. ....... math ka exam end ho gya ..he.......................