Prove the identity cotx*sinx=cosx/(cos^2x+sin^2x)
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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).
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We know that cot x = cos x/sin x
We also know that (sin x)^2 + (cos x)^2 = 1( the fundamental formula of trigonometry).
We'll re-write the identity:
(cos x/sin x)*sin x = cos x/1
We'll simplify by sin x, to the left side and we'll get:
cos x = cos x, true
The identity is proven and cotx*sinx=cosx/(cos^2x+sin^2x), for any measure of the angle x.
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