Question

PROVE THAT :
cot ^2 (secA - 1 ) / 1+ sin A = sec^2 A ( 1 - SinA / 1+ sec A)

Answer 1

Prove :cot square(secA-1)/1+sin=sec square(1-sinA)

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 (Proved)

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Answer 2

Step-by-step explanation:

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