Question

Solve $\dfrac{{\left( {1 - \tan 2^\circ \cot 62^\circ } \right) }}{{\left( {\tan 152^\circ - \cot 88^\circ } \right)}} = $
A. $\sqrt 3 $
B. $ - \sqrt 3 $
C. $\sqrt 2 - 1$
D. $1 - \sqrt 2 $

Answer 1

Hint: To solve this question, we need to observe the given equation and then try to convert all the trigonometric functions into a single function like converting \cotangent into \tangent function after that we simplify it with the help of trigonometric identities and find our solution.

Formula Used:
1. $\tan \left( {90 - \theta } \right) = \cot \theta $
2. $\tan \left( {90 + \theta } \right) = - \cot \theta $
3. $\cot \left( {x + y} \right) = \dfrac{{\cot x\cot y - 1}}{{\cot x + \cot y}}$
4. $\cot \left( {90 + \theta } \right) = - \tan \theta $

Complete step by step solution:
 We need to find the value of $\dfrac{{\left( {1 - \tan 2^\circ \cot 62^\circ } \right) }}{{\left( {\tan 152^\circ - \cot 88^\circ } \right)}}$
Now, we write the given equation as
$\dfrac{{\left( {1 - \tan 2^\circ \cot 62^\circ } \right) }}{{\left( {\tan 152^\circ - \cot 88^\circ } \right)}} = \dfrac{{\left( {1 - \tan \left( {{{90}^\circ} - {{88}^\circ}} \right)\cot {{62}^\circ}} \right)}}{{\left( {\tan \left( {{{90}^\circ} + {{62}^\circ}} \right) - \cot {{88}^\circ}} \right)}}$
Now by applying the formula
$\tan \left( {90 - \theta } \right) = \cot \theta \\ \tan \left( {90 + \theta } \right) = - \cot \theta $
We get
$\dfrac{{\left( {1 - \tan \left( {{{90}^\circ} - {{88}^\circ}} \right)\cot {{62}^\circ}} \right)}}{{\left( {\tan \left( {{{90}^\circ} + {{62}^\circ}} \right) - \cot {{88}^\circ}} \right)}} = \dfrac{{1 - \cot {{88}^\circ}\cot {{62}^\circ}}}{{\cot {{62}^\circ}\cot {{88}^\circ}}}$
Now we apply the formula $\cot \left( {x + y} \right) = \dfrac{{\cot x\cot y - 1}}{{\cot x + \cot y}}$ in the above equation, we get
$\dfrac{{1 - \cot {{88}^\circ}\cot {{62}^\circ}}}{{\cot {{62}^\circ}\cot {{88}^\circ}}} = \cot \left( {{{88}^\circ} + {{62}^\circ}} \right)$
Further solving, we get
$\cot \left( {{{88}^\circ} + {{62}^\circ}} \right) = \cot \left( {{{150}^\circ}} \right)$
$\cot \left( {{{150}^\circ}} \right) = \cot \left( {{{90}^\circ} + {{60}^\circ}} \right)$
Now by applying the formula $\cot \left( {90 + \theta } \right) = - \tan \theta $, we get
$\cot \left( {{{90}^\circ} + {{60}^\circ}} \right) = - \tan {60^\circ}$
(We know that $\tan {60^\circ} = \sqrt 3 $)
So, $ - \tan {60^\circ} = \sqrt 3 $ or $\tan {60^\circ} = - \sqrt 3 $
Therefore, the solution of $\dfrac{{\left( {1 - \tan 2^\circ \cot 62^\circ } \right) }}{{\left( {\tan 152^\circ - \cot 88^\circ } \right)}}$is $ - \sqrt 3 $.

Option ‘B’ is correct

Note: The interesting fact here is to convert $\tan$ to $\cot$ by using a complementary rule that is $\tan (90 – \theta) = \cot \theta$ and $\tan (90 + \theta) = − \cot \theta$ later use the expansion of $\cot (A \pm B)$.