SOLUTION: (i) Show that [(cosA+sinA)^2]/[sec^2(A)+2tanA]=cos^2(A) (ii) Hence find all values of A, where 0 < A < 2pie, which satisfy the equation [sec^2(A)+2tanA]/[cosA+sinA)^2]=2(2+tanA).

Algebra ->  Trigonometry-basics -> SOLUTION: (i) Show that [(cosA+sinA)^2]/[sec^2(A)+2tanA]=cos^2(A) (ii) Hence find all values of A, where 0 < A < 2pie, which satisfy the equation [sec^2(A)+2tanA]/[cosA+sinA)^2]=2(2+tanA).       Log On


   

Question 1096482: (i) Show that [(cosA+sinA)^2]/[sec^2(A)+2tanA]=cos^2(A)
(ii) Hence find all values of A, where 0 < A < 2pie, which satisfy the equation [sec^2(A)+2tanA]/[cosA+sinA)^2]=2(2+tanA).

Answer by sandeepvijay(8) About Me  (Show Source):
You can put this solution on YOUR website!
Answer to (i)
L.H.S = %28cos%28A%29+%2B+sin%28A%29%29%5E2+%2F%28sec%5E2%28A%29%2B2tan%28A%29%29
Denominator of L.H.S = sec%5E2%28A%29%2B2tan%28A%29
Since sec%5E2%28A%29+=+1%2B+tan%5E2%28A%29
Denominator of L.H.S = 1%2Btan%5E2%28A%29%2B2tan%28A%29 = %281%2Btan%28A%29%29%5E2 ----------(1)
Converting tanA to sinA and cosA in (1), denominator of L.H.S. becomes %28sin%28A%29%2Bcos%28A%29%29%5E2+%2F+cos%5E2%28A%29
Therefore, L.H.S. = cos%5E2%28A%29+%2A+%28%28cos%28A%29%2Bsin%28A%29%29%2F%28cos%28A%29%2Bsin%28A%29%29%29%5E2 = cos%5E2%28A%29 = R.H.S
Hence Proved
Answer to (ii)
Based on (i) the equation in (ii) can be reduced to 1%2Fcos%5E2%28A%29+=+2%282+%2B+tan%28A%29%29
=> sec%5E2%28A%29+=+4+%2B2tan%28A%29 => 1%2Btan%5E2%28A%29+=+4%2B2tan%28A%29 => tan%5E2%28A%29+-2tan%28A%29+-3+=+0
This is a quadratic equation in tan(A) as variable and can be solved to get
tan%28A%29+=+3 and tan%28A%29+=+-1
Therefore, A = 71 degree, 251 degree, 135 degree and 315 degree