Question

Find the value of $\tan {75}^0$.

Answer 1

Expand the value of $\tan {75}^0$ in some standard angles.

$\tan {75}^0=\tan \left({45}^0+{30}^0\right)$

Expand the above expression using identity $\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$.

$$\begin{gathered} <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>\tan 0^0=0 \\ <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>\tan {30}^0=\frac{1}{\sqrt{3}} \\ <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>\tan {45}^0=1 \\ <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>\tan {60}^0=\sqrt{3} \\ <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>\tan {90}^0=\infty \end{gathered}$$

So, $\tan {75}^0=\tan \left({45}^0+{30}^0\right)$.

$\begin{array}{rcl}\tan {75}^0& =& \frac{\tan {45}^0+\tan {30}^0}{1-\tan {45}^0\tan {30}^0}\\ & =& \frac{1+{\displaystyle \frac{1}{\sqrt{3}}}}{1-\frac{1}{\sqrt{3}}}\\ & =& \frac{\sqrt{3}+1}{\sqrt{3}-1}\end{array}$

Hence, the required final answer is $\frac{\sqrt{3}+1}{\sqrt{3}-1}$.