Question

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Prove
1+cosA/1-cosA= tan^2A/(secA-1)^2

Answer 1

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 

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Step-by-step explanation:

Answer 2

Answer:

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