CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Identities

Prove that ...

Question

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

Open in App

Solution

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

Suggest Corrections

thumbs-up

0

Similar questions

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ}

= \frac{2}{{sin}^{2} θ - {cos}^{2} θ}

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Trigonometric Identities

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Trigonometric Identities

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon