Question

prove that : sintheta - cos theta + 1 / sin theta+ cos theta - 1 = 1 / sec theta - tan theta​

Answer 1

(sinθ - cosθ +1 )/(sinθ +cosθ -1)

dividing numerator and denominator by cosθ

[(sinθ - cosθ +1 )cosθ]/[(sinθ +cosθ -1)/cosθ]

=(tanθ -1 + secθ )/(tanθ +1 - sec θ)

=(tanθ + secθ -1)/(tanθ - sec θ+1)

As, sec²θ- tan²θ = 1

(secθ -tanθ)(secθ +tanθ) = 1

putting this in numerator,

[(tanθ + secθ -(sec²θ- tan²θ)]/(tanθ - sec θ+1)

=[(tanθ + secθ) -(secθ- tanθ)(secθ+tanθ)]/(tanθ - sec θ+1)

=(tanθ+secθ)[1- (secθ - tanθ)]/(tanθ - sec θ+1)

=(tanθ+secθ)[1- secθ + tanθ)]/(tanθ - sec θ+1)

=(tanθ+secθ)

Now, multiplying and dividing by (secθ- tanθ)

[(tanθ+secθ)×(secθ- tanθ)]/(secθ- tanθ)

=(sec²θ- tan²θ)/(secθ- tanθ)

= 1/(secθ- tanθ) 

=RHS

please mark it as brainlist answer...