Math, asked by MRajesh007, 1 year ago

please solve this paper​

Attachments:

Answers

Answered by yashula
0

HEYA!?

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)

Hope it works!!!

Answered by yashshaw91
0

HOPE U WILL BE SATISFIED WITH THE ANSWER...

Attachments:
Similar questions