Question

tan 15° is equivalent to

Answer 1

given is Tan 15°
  
we know that 45-30=15
applying tan on both sides,
   
   Tan(45-30)=Tan 15
  we use Tan(A-B)=tanA-tanB/1+tanAtanB
      
 Tan 15= tan 45 -tan 30/1+tan45tan30
            = 1-1/√3 / 1+1/√3
           =√3-1/√3+1

∴ Tan 15°=√3-1/√3+1

Answer 2

To find: The value of tan 15°

Step-by-step explanation:

Now as we know

$\tan (x-y)=\dfrac{\tan x - \tan y }{1+\tan x \tan y }$

Now tan 15° can be written as tan (45°-30°)

$\tan (45^\circ-30^\circ)= \dfrac{\tan 45^\circ - \tan 30 ^\circ }{1+ \tan 45^\circ \tan 30^\circ}$

Substituting the trigonometric values of tan 45° and tan 30° we have

$\tan (45^\circ-30^\circ)= \dfrac{1- \dfrac{1}{\sqrt{3} } }{1+ 1\times \dfrac{1}{\sqrt{3} }}=\dfrac{\sqrt{3} -1}{\sqrt{3} +1}$

By rationalizing we get

$\tan (45^\circ-30^\circ)=\dfrac{\sqrt{3} -1}{\sqrt{3} +1}\times \dfrac{\sqrt{3} -1}{\sqrt{3} -1}= \dfrac{(\sqrt{3} -1)^2}{(\sqrt{3} )^2-(1)^2} = \dfrac{3+1-2\sqrt{3} }{2} =2-\sqrt{3}$

Hence the value of tan 15° is  2-√3