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Answer»

(1+cot A-cosec A).(1+tanA+secA)= 2

L.H.S.

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

=(sinA+cosA-1)×(cosA+sinA+1)/sinA.cosA

=[(sinA+cosA)^2-(1)^2]/sinA.cosA.

=(sin^2A+cos^2A+2.sinA.cosA-1)/sinA.cosA.

=( 1+2.sinA.cosA -1)/sinA.cosA.

= 2.sinA.cosA/sinA.cosA

= 2 , proved.



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