Question

The value of $\tan {75}^{\circ }-\cot {75}^{\circ }=$

• A. $2\sqrt{3}$
• B. $3\sqrt{2}$
• C. $3$
• D. $1$

Answer 1

The correct option is A $2\sqrt{3}$

$\tan 75°-\cot 75°$

$\Rightarrow$ Now, $\tan 75°=\tan (45°+30°)$

$=\frac{\tan 45°+\tan 30°}{1-\tan 45°\tan 30°}[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}]$

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

$\Rightarrow \tan 75°=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$\cot 75°=\frac{1}{\tan 75°}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

So, $\tan 75°-\cot 75°=$$\frac{\sqrt{3}+1}{\sqrt{3}-1}$$-\frac{\sqrt{3}-1}{\sqrt{3}+1}$

$=\frac{{(\sqrt{3}+1)}^2-{(\sqrt{3}-1)}^2}{(\sqrt{3}-1)(\sqrt{3}+1)}$

$=\frac{3+1+2\sqrt{3}-3-1+2\sqrt{3}}{3-1}$

$=\frac{4\sqrt{3}}{2}$

$=2\sqrt{3}$

Hence, the answer is $2\sqrt{3}.$