I can't do this equation! Prove that (cot⁡θcos⁡θ)/(cot⁡θ+cos⁡θ) = (cos⁡θ)/(1+sin⁡θ)

I'm doing revision for school but I'm stuck on this question. I can't figure it out at all. I think it has something to do with trigonometric identities. Maybe someone else will know what to do?

Prove that (cot⁡θcos⁡θ)/(cot⁡θ+cos⁡θ) = (cos⁡θ)/(1+sin⁡θ)

Thanks!

(cot⁡θcos⁡θ)/(cot⁡θ+cos⁡c) = ((cos⁡θ/sin⁡θ)*cos⁡θ) / (cos⁡θ/sin⁡θ+cos⁡θ)

=> [cos⁡^2θ/sin⁡θ] / [(cos⁡θ + cosθ*sin⁡θ)/sin⁡θ)]

next you cancel out sinθ,

=> cos⁡^2θ / (cos⁡θ + cosθ*sin⁡θ)

then you further cancel cos⁡θ

=> cos⁡θ / (1 + sin⁡θ)

Factor out cotθ is way simpler.

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(cot⁡θcos⁡θ)/(cot⁡θ+cos⁡θ) = (cos⁡θ)/(1+sin⁡θ)

L.H.S. = (cot⁡θcos⁡θ)/(cot⁡θ+cos⁡θ)
L.H.S. = [cosθ*cosθ/sinθ] / (cosθ/sinθ + cosθ)
L.H.S. = [(cosθ)^2 / sinθ] / [cosθ(1+sinθ)/sinθ]

L.H.S. = cosθ / (1 + sinθ) = R.H.S. >=========< Q . E . D.

x = (cotθcosθ) / cosθ+cotθ

1/x = cosθ+cotθ / cosθcotθ

1/x = cosθ/cosθcotθ + cotθ/cosθcotθ

1/x = 1/cosθ + 1/cotθ

1/x = 1/cosθ + sinθ/cosθ

1/x = 1+sinθ / cosθ

x = cosθ/(1+sinθ)

(cot⁡θcos⁡θ)/(cot⁡θ+cos⁡θ)
= (cot⁡θcos⁡θ)/(cot⁡θ(1+sin⁡θ))
= (cos⁡θ)/(1+sin⁡θ)

And ⁡ is supposed to mean something to us?