Question

prove that

1+secQ-tanQ/1+secQ+tanQ = 1-sinQ/cosQ

Answer 1

Question:-to prove

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

Proof:-Take LHS

Put

1=sec^2a-tan^2 in numerator

$\frac{ {sec}^{2} \alpha - {tan}^{2} \alpha + sec \alpha - tan \alpha }{1 + sec \alpha + tan \alpha } \\ \frac{(sec \alpha - tan \alpha )(sec \alpha + tan \alpha ) + (sec \alpha - tan \alpha )}{1 + sec \alpha + tan \alpha } \\ taking \: (sec \alpha - tan \alpha ) \: as \: commom \: from \: numerator \\ \frac{(sec \alpha - tan \alpha )(1 + sec \alpha + tan \alpha )}{(1 + sec \alpha + tan \alpha )} \\ (1 + sec \alpha + tan \alpha \:) \: will \: be \: cancelled \: out \\ = sec \alpha - tan \alpha \\ now \: convert \: \: seca \: \alpha \: and \: tan \alpha \: into \: sin \alpha \: and \: cos \alpha \\ \frac{1}{cos \alpha } - \frac{sin \alpha }{cos \alpha } \\ = \frac{1 - sin \alpha }{cos \alpha }$

which is eqaul to RHS

{hope it helps you}