Using the identity tan2α = \(\rm\frac{2 tan \alpha}{1-tan^{2}\alpha}\), find the value of tan 15°, correct to three decimal places.

[Use √3 = 1.732]

Given

 tan2α = \(\rm\frac{2 tan \alpha}{1-tan^{2}\alpha}\)

Formula used:

Tan 15° = Tan(45 – 30)°

By the trigonometry formula, we know,

Tan (A – B) = (Tan A – Tan B) /(1 + Tan A Tan B)

Calculation

Therefore, we can write,

tan(45 – 30)° = tan 45° – tan 30°/1+ tan 45° tan 30°

Now putting the values of tan 45° and tan 30° from the table we get;

tan(45 – 30)° = (1 – 1/√3)/ (1 + 1.1/√3)

tan (15°) = √3 – 1/ √3 + 1

= (√3 – 1)²/ [(√3)² - 1²]

= (3 + 1 - 2√3)/2 = 2 - √3

= 0.268

Hence, the value of tan (15°) is 0.268.
Alternate Method 
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tan 30° = tan 2(15°)

By the trigonometry formula, we know,

 tan2α = \(\rm\frac{2 tan \alpha}{1-tan^{2}\alpha}\),

Calculation

Therefore, we can write,

tan 30° = 2 × tan 15° /(1 - tan2 15°)

Now putting the values of tan 30° we get;

⇒ 1/ √3 = 2 tan 15° / (1 - tan2 15°)

let tan (15°) = x

⇒ 1/ √3 = 2x / (1 - x2)

⇒ x2 - 1 + 2√3 x = 0

⇒ x2 + 2√3 x - 1 = 0

From quadratic formula,

x = \(\frac{-2√3 \pm \sqrt{(2√3)^2 -4(1)(-1) }}{2\times 1}\)

⇒ x = \(\frac{-2√3 \pm \sqrt{12 +4 }}{2\times 1}\)

⇒ x = \(\frac{-2√3 \pm \sqrt{16 }}{2\times 1}\)

⇒ (4 - 2√3)/2 = 2 - √3

= 0.268

Hence, the value of tan (15°) is 0.268.