Question

The value of $\cot 105°$ is

• A. $\sqrt{3}-2$
• B. $2-\sqrt{3}$
• C. $\sqrt{2}+3$
• D. $\sqrt{3}+2$

Answer 1

The correct option is A

$\sqrt{3}-2$

Find the value of $\cot 105°$:

$\begin{array}{rcl}\cot 105°& =& \cot \left(90°+15°\right)\\ & =& -\tan \left(15°\right)[\because \cot (90°+\mathrm{\theta })=-\mathrm{tan\theta }]\\ & =& -tan\left(45°-30°\right)\\ & =& -\left[\frac{tan\left(45°-tan30°\right)}{1+tan45°tan30°}\right][\because \tan (\mathrm{A}-\mathrm{B})=\frac{\mathrm{tanA}-\mathrm{tanB}}{1+\mathrm{tanAtanB}}]\\ & =& -\left[\frac{1-{\displaystyle \frac{1}{\sqrt{3}}}}{1+1.{\displaystyle \frac{1}{\sqrt{3}}}}\right][\because \tan (45°)=1,\tan (30°)=\frac{1}{\sqrt{3}}]\\ & =& -\left[\frac{\sqrt{3}-1}{\sqrt{3}+1}\right]...\left(1\right)\end{array}$

Now multiply by conjugate of $\left(\sqrt{3}+1\right)$ in equation $\left(1\right)$. then,

$\begin{array}{rcl}\cot 105°& =& -\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)}\times \frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)}\\ & =& -\frac{\left(3-\sqrt{3}-\sqrt{3}+1\right)}{{\left(\sqrt{3}\right)}^2-1^2}[\because (\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})={\mathrm{a}}^2-{\mathrm{b}}^2]\\ & =& \frac{-\left(4-2\sqrt{3}\right)}{2}\\ & =& \frac{-2\left(2-\sqrt{3}\right)}{2}\\ & =& \sqrt{3}-2\end{array}$

Hence, the correct option is A.