Question

Evaluate$\tan 15°$

Answer 1

To evaluate:$\tan 15°$

We can write $15°=45°-30°$

Taking tanget on both the sides, we get

$\tan 15°$ $=$$\tan ({45}^0-{30}^0)$

Applying formula $\tan (\mathrm{A}-\mathrm{B})=\frac{\mathrm{tanB}-\mathrm{tanB}}{1+\mathrm{tanAtanB}}$, we get

$\tan 15°$ $=$$\frac{\tan {45}^0-\tan {30}^0}{1+\tan {45}^0\tan {30}^0}$

$=$ $\frac{1-{\displaystyle \frac{1}{\surd 3}}}{1+1\times \frac{1}{\surd 3}}$ (From trigonometric angle table $\tan {45}^0=1$and $\tan {30}^0=\frac{1}{\sqrt{3}}$)

$=$$\frac{{\displaystyle \frac{\surd 3-1}{\surd 3}}}{1+\frac{1}{\surd 3}}$

$=$ $\frac{{\displaystyle \frac{(\surd 3-1)}{\surd 3}}}{\frac{(\surd 3+1)}{\surd 3}}$

$=$$\frac{(\surd 3-1)}{\surd 3}\times \frac{\surd 3}{(\surd 3+1)}$

$=$$\frac{\surd 3-1}{\surd 3+1}$

Thus, the required value of $\tan 15°$$=$$\frac{\surd 3-1}{\surd 3+1}$.