Math, asked by shivanshuaman9, 7 months ago

Cot^2A(secA-1)/1+sinA=sec^2A*(1-sinA/1+secA. Prove this

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Answered by Anonymous
0

Answer:

CRM

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

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Answered by Brainlyboy00
0

Answer:

plz mark brainlist dear

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