We have to prove that (1+ sin x + cos x) / (1+ sin x - cos x) = cot x/2

Now sin 2x = 2 sin x * cos x and cos 2x = 2 (cos x)^2 - 1

(1+ sin x + cos x) / (1+ sin x - cos x)

=> [1+2*(sin x/2)*(cos x/2) + 2*(cos x/2)^2 - 1]/ [1 + 2*(sin x/2)*(cos x/2)-1+2( sin x/2)^2]

eliminate 1 as we have -1 and +1 in the numerator and denominator.

[2*(sin x/2)*(cos x/2) + 2*(cos x/2)^2]/ [2*(sin x/2)*(cos x/2)+2( sin x/2)^2]

cancel 2 from the numerator and denominator

=> [(sin x/2)*(cos x/2) + (cos x/2)^2 ] / [(sin x/2)*(cos x/2) + ( sin x/2)^2]

factorize

=> [cos x/2*( sin x/2 + cos x/2)]/[sin x/2*( cos x/2 + sin x/2)]

cancel (sin x/2 + cos x/2)

=> (cos x/2) / (sin x/2)

=> cot x/2

Therefore we have proved that **(1+ sin x + cos x) / (1+ sin x - cos
x) = cot x/2**

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