Question 62235
1 + cos A/ 1 - cos A = tanČA/(secA - 1)Č

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

RHS = tanČA/(secA - 1)Č
    = [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

Regards
Kapil
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