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Trigonometric Identities

Prove that ...

Question

Prove that $\frac{s i n θ + s i n 2 θ}{1 + c o s θ + c o s 2 θ} = t a n θ$

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Solution

$(\frac{s i n θ + s i n 2 θ}{1 + c o s θ + c o s 2 θ}) = (\frac{s i n θ + 2 s i n θ c o s θ}{1 + c o s 2 θ + c o s θ}) = (\frac{s i n θ (1 + 2 c o s θ)}{1 + 2 c o s^{2} θ - 1 + c o s θ}) = (\frac{s i n θ (1 + 2 c o s θ)}{c o s θ (1 + 2 c o s θ)}) = t a n θ = R H S$

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