Question

The value of $\tan {75}^{\circ} - \cot {75}^{\circ} $ is-
A. $2\sqrt 3 $
B. $3\sqrt 2 $
C. 3
D. 1

Answer 1

Hint: We will use the formulas for various trigonometric functions which are given below-
$
  \dfrac{{\sin x}}{{\cos x}} = \tan x \\
  \cos 2x = {\cos ^2}x - {\sin ^2}x \\
  \sin 2x = 2\sin x\cos x \\
$

Complete step-by-step solution -
To find the value of the given expression, we will first convert it in terms of sine and cosine functions and then solve further as-
$
   = \dfrac{{\sin {{75}^o}}}{{\cos {{75}^o}}} - \dfrac{{\cos {{75}^o}}}{{\sin {{75}^o}}} \\
   = \dfrac{{{{\sin }^2}{{75}^o} - {{\cos }^2}{{75}^o}}}{{\sin {{75}^o}\cos {{75}^o}}} \\
$
We will multiply and divide the expression by 2-
$ = \dfrac{{ - 2\left( {{{\cos }^2}{{75}^o} - \sin {{75}^o}} \right)}}{{2\sin {{75}^o}\cos {{75}^o}}}$
Applying the formula for cos2x and sin2x,
$
= \dfrac{{ - 2\cos {{150}^{\text{o}}}}}{{\sin {{150}^{\text{o}}}}} \\
= - 2\cot {150^{\text{o}}} = - 2\cot \left( {{{180}^{\text{o}}} - {{30}^{\text{o}}}} \right) \\
Using\;\cot \left( {180 - {\text{x}}} \right) = - cotx \\
= 2\cot {30^{\text{o}}} = 2\sqrt 3 \\
$
This is the required answer. The correct option is A.

Note: Instead of using this method, we can use a more calculative method by writing $75^o$ as $(45 + 30)^o$ and applying the angle difference formula as-
$
   = \tan {\left( {45 + 30} \right)^{\text{o}}} - \dfrac{1}{{\tan {{\left( {45 + 30} \right)}^{\text{o}}}}} \\
   = \dfrac{{\tan {{45}^{\text{o}}} + \tan {{30}^{\text{o}}}}}{{1 - \tan {{45}^{\text{o}}}\tan {{30}^{\text{o}}}}} - \dfrac{{1 - \tan {{45}^{\text{o}}}\tan {{30}^{\text{o}}}}}{{\tan {{45}^{\text{o}}} + \tan {{30}^{\text{o}}}}} \\
  Applying\;the\;values - \\
   = \dfrac{{1 + \dfrac{1}{{\sqrt 3 }}}}{{1 - \dfrac{1}{{\sqrt 3 }}}} - \dfrac{{1 - \dfrac{1}{{\sqrt 3 }}}}{{1 + \dfrac{1}{{\sqrt 3 }}}} = 2\sqrt 3 \\
$