Question

prove that. please somebody do this very very urgent exam please

Answer 1

Answer:

LHS

=cot²A(secA-1)/(1+sinA)

={(cos²A/sin²A)(1/cosA-1)}/(1+sinA)

=[{cos²A/(1-cos²A)}{(1-cosA)/cosA}]/(1+sinA)

=[{cos²A/(1+cosA)(1-cosA)}×{(1-cosA)cosA}]/(1+sinA)

={cosA/(1+cosA)}/(1+sinA)

=cosA/(1+sinA)(1+cosA)

RHS

=sec²A(1-sinA)/(1+secA)

=(1/cos²A)(1-sinA)/(1+1/cosA)

={(1-sinA)/cos²A}/{(1+cosA)/cosA}

={(1-sinA)/(1-sin²A)}/{(1+cosA)/cosA}

={(1-sinA)/(1+sinA)(1-sinA)}/{(1+cosA)/cosA}

={1/(1+sinA)}/{(1+cosA)/cosA}

=cosA/(1+sinA)(1+cosA)

∴, LHS=RHS

take it on either side and you will get your ans equals to 0

Answer 2

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