Math, asked by shiekhbilkees010, 11 months ago

(sinA/1-cosA -1-cosA/sinA)(cosA/1-sinA-1-sinA/cosA)

Answers

Answered by hmf

0

Answer:

1/(1-sinA)(1-cosA)

Step-by-step explanation:

=[Sin2A-(1-cosA)2/(1-cosA)sinA] [cos2A-(1-sinA)2/(1-sinA)cosA]

= sinAcosA/(1-sinA)(1-cosA)sinAcosA

=1/(1-sinA)(1-cosA)

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