Question

prove that sin theta minus cos theta + 1 by sin theta + cos theta minus 1 is equal to 1 by secant theta minus tan theta ​

Answer 1

Answer:

HI BUDDY

LHS:

$\frac{sin \alpha - cos \alpha + 1}{sin \alpha + \cos \alpha - 1 } \: \\ \\ dividing \: numerator \: and \: denominator \: by \: cos \alpha \\ \\ \frac{tan \alpha - 1 + sec \alpha }{tan \alpha + 1 - sec \alpha } \\ \\ \frac{tan \alpha + sec \alpha - 1}{tan \alpha + 1 - sec \alpha } \\ \\ \frac{tan \alpha + sec \alpha - ( {sec}^{2} \alpha - {tan}^{2} \alpha ) }{tan \alpha + 1 - sec \alpha } \: \: \: \: ( {sec}^{2} - {tan}^{2} = 1) \\ \\ \frac{sec \alpha + tan \alpha - ((sec \alpha + tan \alpha )(sec \alpha - tan \alpha )) }{tan \alpha + 1 - sec \alpha } \\ \\ \frac{sec \alpha + tan \alpha (1 - (sec \alpha - tan \alpha ))}{tan \alpha + 1 - sec \alpha } \: \: \: taking \: sec \alpha + tan \alpha \: common \\ \\ \frac{sec \alpha + tan \alpha (tan \alpha + 1 - sec \alpha )}{tan \alpha + 1 -sec \alpha } \\ \\ sec \alpha + tan \alpha - - - - - - - - - 1 \\ \\ now \: we \: know \: that \\ \\ 1 + {tan}^{2} \alpha = {sec}^{2} \alpha \\ \\ {sec}^{2} \alpha - {tan}^{2} \alpha = 1 \\ \\ (sec \alpha + tan \alpha )(sec \alpha - tan \alpha ) = 1 \\ \\ sec \alpha + tan \alpha = \frac{1}{sec \alpha - tan \alpha } - - - - - - - 2 \\ \\ from \: 1 \: and \: 2 \\ \\ \frac{sin \alpha - cos \alpha + 1}{sin \alpha + cos \alpha - 1} = \frac{1}{sec \alpha - tan \alpha } \\ \\ therefore \: lhs = rhs \\ \\ hence \: proved \\ \\ hope \: this \: helps \\ \\ please \: mark \: as \: brainliest$