Question

prove 1+cos A / 1-cos A= tan^2/(secA-1)^2​

Answer 1

PLEASE MARK IT AS A BRAINLIEST AND FOLLOW ME

Answer:

1 + cos A/ 1 - cos A = tan²A/(secA - 1)²

LHS = (1 + cos A)/ (1 - cos A)

RHS = tan²A/(secA - 1)²

= [tanA/(sec-1)]^2

= [tanA/((1/cosA)-1)]^2

= [tanA/(1-cosA)/cosA]^2

= [(tanA*cosA)/(1-cosA)]^2

= [sinA/(1-cosA)]^2

= sinA^2/(1-cosA)^2

= (1-cosA^2)/(1-cosA)^2

= [(1-cosA)*(1+cosA)]/[(1-cosA)*(1-cosA)]

= (1+cosA)/(1-cosA)

LHS = RHS