1+cosA/1-cosA=tan^A/(secA-1)
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recheck the ques
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i guess the ques
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
hence proved
:)hope the ans would help u
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
hence proved
:)hope the ans would help u
thanks a lot...
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