# (sin x/2+cos x/2)^2 - 1/tg x/2-sin x/2cos x/2=2ctg^2 x/2

sciencesolve | Teacher | (Level 3) Educator Emeritus

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The request of the problem is vague, hence, supposing that you need to prove that the given expression holds, you should expand the square such that:

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

Notice that `(sin(x/2))/(cos(x/2))`  substitutes `tan(x/2)`  and `(cos^2(x/2))/(sin^2(x/2))`  substitutes `cot^2(x/2).`

You need to use the fundamental formula of trigonometry such that:

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

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

You should bring the terms to a common denominator such that:

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

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

`sin^2(x/2)cos(x/2) - cos(x/2)sin(x/2) = cos x`

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

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

`(1+cos x)*cos(x/2) = 2cos x + sin x`

Hence, evaluating the given expression yields that it does not represent an identity.

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