Question

tanA-cosecA.secA(1-2cosA)=cotA​

Answer 1

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

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

(1+cot A-cosec A).(1+tanA+secA)= 2L.H.S.=(1+cosA/sinA-1/sinA).(1+sinA/cosA+1/cosA)

(1+cot A-cosec A).(1+tanA+secA)= 2L.H.S.=(1+cosA/sinA-1/sinA).(1+sinA/cosA+1/cosA)=(sinA+cosA-1)×(cosA+sinA+1)/sinA.cosA

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

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

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

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