Prove that (1+cotA-cosecA)(1±tanA+secA)=2
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Answer:
Step-by-step explanation:
(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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