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harina [27]
2 years ago
11

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

Mathematics
1 answer:
Vitek1552 [10]2 years ago
7 0

(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))

Recall that

tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)

so cos²(<em>θ</em>) cancels with the cos²(<em>θ</em>) in the tan²(<em>θ</em>) term:

(sin²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))

Recall the double angle identity for cosine,

cos(2<em>θ</em>) = 2 cos²(<em>θ</em>) - 1

so the 1 in the denominator also vanishes:

(sin²(<em>θ</em>) - 1) / (2 cos²(<em>θ</em>))

Recall the Pythagorean identity,

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

which means

sin²(<em>θ</em>) - 1 = -cos²(<em>θ</em>):

-cos²(<em>θ</em>) / (2 cos²(<em>θ</em>))

Cancel the cos²(<em>θ</em>) terms to end up with

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

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