**Prove the identity cotx*sinx=cosx/(cos^2x+sin^2x)**

We have to prove that cot x*sin x = cos x /((cos x)^2 + (sin x)^2)

Now we know that (cos x)^2 + (sin x)^2 = 1

Also, cot x = cos x / sin x

So cot x*sin x = (cos x / sin x)* sin x = cos x

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

Therefore both the sides are equal to cos x.

**We prove that cot x*sin x = cos x /((cos x)^2 + (sin
x)^2).**

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

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**

**Prove the following identity: cos x + cos 2x + cos 3x = cos 2x(1 + 2cos x)**

We have to prove : cos x + cos 2x + cos 3x = cos 2x(1 + 2cos x)

We know that cos A + cos B = 2 * cos (A + B)/2 * cos (A − B)/2

Now cos x + cos 2x + cos 3x

=> 2 * cos (4x / 2) * cos (2x / 2) + cos 2x

=> 2 cos 2x * cos x + cos 2x

=> cos 2x ( 1 + 2 cos x)

Therefore we prove that

**cos x + cos 2x + cos 3x = cos 2x(1 + 2cos x)**

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