Question

# Find the value of $\left( {\cot {9^ \circ } - \cot {{27}^ \circ } - \cot {{63}^ \circ } + \cot {{81}^ \circ }} \right)$:(A) 4(B) 0(C) 3(D) None of these

Hint: Use $\cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta$, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ to convert the complete expression in $\sin \theta$ and $\cos \theta$. And then use standard results to find the final value.

The given expression is $\left( {\cot {9^ \circ } - \cot {{27}^ \circ } - \cot {{63}^ \circ } + \cot {{81}^ \circ }} \right)$. Let its value is $x$. Then we have:
$x = \left( {\cot {9^ \circ } - \cot {{27}^ \circ } - \cot {{63}^ \circ } + \cot {{81}^ \circ }} \right)$
Now, we know that $\cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta$. Using this, we have $\cot {81^ \circ } = \tan {9^ \circ }$ and $\cot {63^ \circ } = \tan {27^ \circ }$. Putting these values, weâ€™ll get:
$\Rightarrow x = \cot {9^ \circ } - \cot {27^ \circ } - \tan {27^ \circ } + \tan {9^ \circ }, \\ \Rightarrow x = \tan {9^ \circ } + \cot {9^ \circ } - \left( {\tan {{27}^ \circ } + \cot {{27}^ \circ }} \right) \\$
Using $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$, weâ€™ll get:
$\Rightarrow x = \dfrac{{\sin {9^ \circ }}}{{\cos {9^ \circ }}} + \dfrac{{\cos {9^ \circ }}}{{\sin {9^ \circ }}} - \left( {\dfrac{{\sin {{27}^ \circ }}}{{\cos {9^ \circ }}} + \dfrac{{\cos {9^ \circ }}}{{\sin {9^ \circ }}}} \right), \\ \Rightarrow x = \dfrac{{{{\sin }^2}{9^ \circ } + {{\cos }^2}{9^ \circ }}}{{\sin {9^ \circ }\cos {9^ \circ }}} - \left( {\dfrac{{{{\sin }^2}{{27}^ \circ } + {{\cos }^2}{9^ \circ }}}{{\sin {{27}^ \circ }\cos {{27}^ \circ }}}} \right), \\ \Rightarrow x = \dfrac{1}{{\sin {9^ \circ }\cos {9^ \circ }}} - \dfrac{1}{{\sin {{27}^ \circ }\cos {{27}^ \circ }}}, \\ \Rightarrow x = \dfrac{2}{{2\sin {9^ \circ }\cos {9^ \circ }}} - \dfrac{2}{{2\sin {{27}^ \circ }\cos {{27}^ \circ }}} \\ \$
We know that $2\sin \theta \cos \theta = \sin 2\theta$.Using this weâ€™ll get:
$\Rightarrow x = \dfrac{2}{{\sin {{18}^ \circ }}} - \dfrac{2}{{\sin {{54}^ \circ }}}$
If we use $\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta$, weâ€™ll get $\sin {54^ \circ } = \cos {36^ \circ }$. Putting this weâ€™ll get:
$\Rightarrow x = \dfrac{2}{{\sin {{18}^ \circ }}} - \dfrac{2}{{\cos {{36}^ \circ }}}$
We know that $\sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}$ and $\cos {36^ \circ } = \dfrac{{\sqrt 5 + 1}}{4}$. Putting these values, we have:
$\Rightarrow x = \dfrac{2}{{\dfrac{{\sqrt 5 - 1}}{4}}} - \dfrac{2}{{\dfrac{{\sqrt 5 + 1}}{4}}}, \\ \Rightarrow x = \dfrac{8}{{\sqrt 5 - 1}} - \dfrac{8}{{\sqrt 5 + 1}}, \\ \Rightarrow x = 8\left( {\dfrac{{\sqrt 5 + 1 - \sqrt 5 + 1}}{{{{\left( {\sqrt 5 } \right)}^2} - {1^2}}}} \right), \\ \Rightarrow x = 8 \times \dfrac{2}{4}, \\ \Rightarrow x = 4 \\$
Thus, the value of the expression is 4. (A) is the correct option.

Note: If in $\sin \theta$ or $\cos \theta$, $\theta$ is a multiple of 6, we can determine the values using formulae such as:
$\sin 2\theta = 2\sin \theta \cos \theta , \\ \cos 2\theta = 2{\cos ^2}\theta - 1 \\ \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B, \\ \sin ({90^ \circ } - \theta ) = \cos \theta \\$