Math, asked by sureaKumar, 1 year ago

what is its solution ​

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Answered by sudeepthichunduru
0

I did not understand question properly

Answered by rakhithakur
0
cosA-sinA+1/cosA+sinA-1
=(cosA-sinA+1)(cosA+sinA+1)/(cosA+sinA-1)(cosA+sinA+1)

=(cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²}

=(cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1)

={cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1]

=(cos²A+2cosA+cos²A)/2cosAsinA

=(2cos²A+2cosA)/2cosAsinA

=2cosA(cosA+1)/2cosAsinA

=(cosA+1)/sinA

=cosA/sinA+1/sinA

=cotA+cosecA


=cosecA+cotA (Proved)

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