Prove that

Question

[1+cotA-CosecA] [1+tanA+secA] = 2

Answer 1

[1+cotA-CosecA]*[1+tanA+secA]
= 1 + tanA + secA + cotA + 1 + cosecA - (1/sinA) - (1/cosA) - secAcosecA
= 1 + tanA + secA + cotA + 1 + cosecA - cosecA - secA - secAcosecA
= 2 + tanA + cotA - secAcosecA
= 2 + [(sin^2A + cos^2A)/(sinAcosA)] - secAcosecA
= 2 + 1/(sinAcosA) - secAcosecA
= 2 + secAcosecA - secAcosecA
= 2

Answer 2

1+cot-csc = (sin+cos-1)/sin
1+tan+sec = (sin+cos+1)/cos
multiply them and you have
((sin+cos)^2-1)/(sin*cos)
= (1+2sin*cos-1)/(sin*cos)
= 2