Question

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Answer 1

Answer:

Step-by-step explanation:

LHS=(sin0-cos0+1) / (sin0+cos0-1)

dividing by cos0 with both numerator and denominator

=,(sin0/cos0-cos0/cos0+1,/cos0) / (sin0/cos0+cos0/cos0-1/cos0

=(tan0+ sec0-1) / (tan0-sec0+1)

multiply (tan0-sec0) with both numerator and denominator

=(tan0+sec0-1) ( tan0-sec0) / (tan0-sec0+1) (tan0-sec0)

=[(tan^0-sec^0) - ( tan0-sec0)] / (tan0-sec0+1)(tan0-sec0)

=(-1-tan0+sec0) / (tan0-sec0+1)( tan0- sec0). [hence, sec^0-tan^0= 1]

= -1/(tan0-sec0)

= 1 / ( sec0-tan0)= RHS

Answer 2

(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