Question

Find the value of $\tan {15}^{\circ }$.

Answer 1

Find the value of given trigonometric ratio by using the trigonometric identity

Given trigonometric ratio: $\tan {15}^{\circ }$

$\tan {15}^{\circ }$can be expressed as, $\tan {15}^{\circ }=\tan ({45}^{\circ }-{30}^{\circ })$

We know that $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

So by applying the above identity we get,

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

As, $\tan {45}^{\circ }=1,$ and $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$.

By substituting the values we get,

$$\begin{gathered} \Rightarrow \tan {15}^{\circ }=\frac{1-\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}{1+\left(1\right)\left(\frac{1}{\sqrt{3}}\right)} \\ =\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} \\ =\frac{\sqrt{3}-1}{\cancel{\sqrt{3}}}\times \frac{\cancel{\sqrt{3}}}{\sqrt{3}+1} \\ =\frac{\sqrt{3}-1}{\sqrt{3}+1} \end{gathered}$$

Hence, the value of $\tan {15}^{\circ }$is $\frac{\sqrt{3}-1}{\sqrt{3}+1}$.