Prove that (1+cos A- cosec A) (1+tan A+secA) = 2
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(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.
Answered by
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HEYA!!!
(1+Cot A - Cosec A) (1+TanA+SecA)
= (sinA + cosA - 1)/sinA (sinA + cosA +1)/cosA
= [(sinA + cosA)^2 - 1] /sinAcosA
= (sinA^2 + cosA^2 + 2sinAcosA - 1)/sinAcosA
= (1-1 + 2sinacosA) /sinAcosA = 2
......HOPE IT HELPS........
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