Question 1096482
Answer to (i)

L.H.S = {{{(cos(A) + sin(A))^2 /(sec^2(A)+2tan(A))}}}

Denominator of L.H.S = {{{sec^2(A)+2tan(A)}}}

Since {{{sec^2(A) = 1+ tan^2(A)}}}

Denominator of L.H.S = {{{1+tan^2(A)+2tan(A)}}} = {{{(1+tan(A))^2}}} ----------(1)

Converting tanA to sinA and cosA in (1), denominator of L.H.S. becomes {{{(sin(A)+cos(A))^2 / cos^2(A)}}}

Therefore, L.H.S. = {{{cos^2(A) * ((cos(A)+sin(A))/(cos(A)+sin(A)))^2}}} = {{{cos^2(A)}}} = R.H.S

Hence Proved

Answer to (ii)

Based on (i) the equation in (ii) can be reduced to {{{1/cos^2(A) = 2(2 + tan(A))}}}

=> {{{sec^2(A) = 4 +2tan(A)}}} => {{{1+tan^2(A) = 4+2tan(A)}}} => {{{tan^2(A) -2tan(A) -3 = 0}}}

This is a quadratic equation in tan(A) as variable and can be solved to get
{{{tan(A) = 3}}} and {{{tan(A) = -1}}}

Therefore, A = 71 degree, 251 degree, 135 degree and 315 degree