Using the identity tan2a =(2tanα)/(1-tan²α) , find the value of tan15, and correct it to three decimal places. [Use √3 = 1.732]

Using the identity tan2a =(2tanα)/(1-tan²α) , find the…

Using the identity tan2a $=\frac{2tan\alpha}{1-tan^2\alpha}$ , find the value of tan15, and correct it to three decimal places.

[Use √3 = 1.732]

Show answer & explanation

Correct answer: A

Explanation

The correct answer is Option 1 i.e. 0.268.

The given identity = tan2a $=\frac{2tan\alpha}{1-tan^2\alpha}$

tan 30° = 2 × tan 15°/(1 - tan²15°)

let tan 15° = x

Now, tan 30 = 1/√3

⇒ 1/√3 = 2x/(1 - x²)

⇒ (1 - x²) = 2√3x

⇒ x² + 2√3 - 1 = 0

By using quadratic formulae

⇒ x = [-b ± √(b² - 4ac)]/2a

⇒ x = (-2√3 ± √(((2√3)² - 4ac)]/2 × 1

⇒ x = (-2√3 ± √16)/2

⇒ x = (4 - 2√3)/2

⇒ 2 - √3

⇒ 2 - 1.732 = 0.268