Question

Prove that sin theta - 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

Answer 2

Answer:

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