Question

Evaluate tan 15°without using calculator

Answer 1

Solution

⇒We have to find the value tan15°

Now we can write as

⇒tan(45° - 30°) = tan15°

Using this formula

⇒tan(A - B) = (tanA - tanB)/(1 + tanAtanB)

we get

⇒tan(45° - 30°) = (tan45° - tan30°)/(1 + tan45°tan30°)

We know that

⇒tan45° = 1

⇒tan30° = 1/√3

we get

⇒tan(45° - 30°) = (1 - 1/√3)/(1 + 1×1/√3)

⇒tan(45° - 30°) = [( √3 - 1)/√3]/[(√3 + 1)/√3]

⇒tan(45° - 30°) = ( √3 - 1)/√3 × √3/(√3 + 1)

⇒tan(45° - 30°) = (√3 - 1)/(√3 + 1)

Answer

⇒tan15° =  (√3 - 1)/(√3 + 1)

Answer 2

Step-by-step explanation:

$✯\begin{gathered} \frac{1 - { \tan(15) }^{2} }{1 + { \tan(15) }^{2} } \\ \\ \pink{ \cos(2 \alpha ) = \frac{1 - { \tan( \alpha ) }^{2} }{1 + { \tan( \alpha ) }^{2} } } \\ \\ \blue{hence} \\ \\ \frac{1 - { \tan(15) }^{2} }{1 + { \tan(15) }^{2} } = \cos(2 \times 15) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \cos(30) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ \sqrt{3} }{2} \\ \\ \green{\frac{1 - { \tan(15) }^{2} }{1 + { \tan(15) }^{2} } = \frac{ \sqrt{3} }{2} }\end{gathered}$