Question

prove that (1+cotA-cosecA) (1+tanA+ secA)=2

Answer 1

Convert the left side in terms of sines and cosines to get: 
LHS = [1 + cot(A) - csc(A)][1 + tan(A) + sec(A)] 
= [1 + cos(A)/sin(A) - 1/sin(A)][1 + sin(A)/cos(A) + 1/cos(A)] 
= [sin(A) + cos(A) - 1]/sin(A) * [sin(A) + cos(A) + 1]/cos(A), by getting common denominators 
= {[sin(A) + cos(A)]^2 - 1}/[sin(A)cos(A)], via difference of squares 
= [sin^2(A) + 2sin(A)cos(A) + cos^2(A) - 1]/[sin(A)cos(A)], by expanding 
= [2sin(A)cos(A) + 1 - 1]/[sin(A)cos(A)], since sin^2(A) + cos^2(A) = 1 
= [2sin(A)cos(A)]/[sin(A)cos(A)] 
= 2, by canceling sin(A)cos(A) 
= RHS. 

Answer 2

here is the answer hope it helps