Question

prove that
sinA - cosA +1 / sinA + cosA -1 = 1/ secA - tanA​

Answer 1

Answer:

step-by-step explanation:

Given,

L.H.S

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

Dividing both numerator and denominator by cos θ ,

we get,

= $\frac{sin θ/cos θ - cos θ/cos θ +1/cos θ}{sin θ/cos θ + cos θ/cos θ -1/cos θ}$

= $\frac{tanθ - 1 + secθ}{tan θ+ sec θ -1 }$

Now,

multiply both the numerator and denominator by ( tan θ - sec θ )

we get,

= $\frac{(tanθ - 1 + secθ)/( tan θ-secθ )}{(tan θ+ sec θ -1)( tan θ - sec θ )}$

=

$\frac{[tex]{tan}^{2}$θ - ${secθ}^{2}$θ -(tan θ - sec θ)}{(tan θ- sec θ +1)( tan θ - sec θ )}[/tex]

= $\frac{-tanθ - 1 + secθ}{(tan θ- sec θ +1)( tan θ - sec θ )}$

( °.° ${sec}^{2}$θ - ${tan}^{2}$θ = 1 )

= $\frac{-1}{tan θ- sec θ}$

= $\frac{1}{sec θ-tan θ}$

= R.H.S.

Hence , proved.