(\tan ^(2)\theta \cos ^(2)\theta -1)/(1+\cos (2\theta ))=

(tan²(θ) cos²(θ) - 1) / (1 + cos(2θ))

Recall that

tan(θ) = sin(θ) / cos(θ)

so cos²(θ) cancels with the cos²(θ) in the tan²(θ) term:

(sin²(θ) - 1) / (1 + cos(2θ))

Recall the double angle identity for cosine,

cos(2θ) = 2 cos²(θ) - 1

so the 1 in the denominator also vanishes:

(sin²(θ) - 1) / (2 cos²(θ))

Recall the Pythagorean identity,

cos²(θ) + sin²(θ) = 1

which means

sin²(θ) - 1 = -cos²(θ):

-cos²(θ) / (2 cos²(θ))

Cancel the cos²(θ) terms to end up with

(tan²(θ) cos²(θ) - 1) / (1 + cos(2θ)) = -1/2

idk

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