# Prove the following identity: (1+ sin x + cos x) / (1+ sin x - cos x) = cot x/2

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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 andÂ +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**

We know that 1 + cos x = 2(cos x/2)^2 (1)

We'll write sin x = sin 2(x/2) = 2sin(x/2)*cos (x/2) (2)

We'll add (1) + (2):

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

We'll factorize by 2cos (x/2):

1+ sin x + cos x = 2cos (x/2)[cos (x/2) + sin(x/2)]

We'll write the terms from denominator as:

1 - cos x = 2(sin x/2)^2 (3)

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

We'll add (3) + (2):

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

We'll factorize by 2sin (x/2):

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

We'll re-write the numerator and denominator of the ratio:

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

We'll simplify the brackets:

(1+ sin x + cos x) / (1+ sin x - cos x) = 2cos (x/2)/2sin(x/2)

We'll simplify and we notice that the ratio cos (x/2)/sin(x/2) = cot (x/2)

**(1+ sin x + cos x) / (1+ sin x - cos x) = cot (x/2) q.e.d.**