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

Answer:

Step-by-step explanation:

here is your answer

look and pay extra attention in step 3

i used here the identitya^2-b^3