Question

Find the value of $\dfrac{{{{\cot }^2}{{15}^ \circ } - 1}}{{{{\cot }^2}{{15}^ \circ } + 1}}$.
A. $\dfrac{1}{2}$
B. $\dfrac{{\sqrt 3 }}{2}$
C. $\dfrac{{3\sqrt 3 }}{4}$
D. $\sqrt 3 $

Answer 1

Hint: First convert the given expression in $\tan $ function. Then by using double angle identities we will simplify the given expression. Then with the help of the trigonometry table, we will calculate the value of the expression.

Formula Used:
$\cot \theta = \dfrac{1}{{\tan \theta }}$
$\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$

Complete step by step solution:
Given expression is
$\dfrac{{{{\cot }^2}{{15}^ \circ } - 1}}{{{{\cot }^2}{{15}^ \circ } + 1}}$
Now we apply $\cot \theta = \dfrac{1}{{\tan \theta }}$ in the given expression
$ = \dfrac{{\dfrac{1}{{{{\tan }^2}{{15}^ \circ }}} - 1}}{{\dfrac{1}{{{{\tan }^2}{{15}^ \circ }}} + 1}}$
Simplify the expression
$ = \dfrac{{\dfrac{{1 - {{\tan }^2}{{15}^ \circ }}}{{{{\tan }^2}{{15}^ \circ }}}}}{{\dfrac{{1 + {{\tan }^2}{{15}^ \circ }}}{{{{\tan }^2}{{15}^ \circ }}}}}$
$ = \dfrac{{1 - {{\tan }^2}{{15}^ \circ }}}{{1 + {{\tan }^2}{{15}^ \circ }}}$
Now we will apply the formula $\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$
$ = \cos \left( {2 \cdot {{15}^ \circ }} \right)$
$ = \cos \left( {{{30}^ \circ }} \right)$
Now putting the value of $\cos {30^ \circ }$
$ = \dfrac{{\sqrt 3 }}{2}$

Option ‘B’ is correct

Additional Information:
The double angle formula that is used to express 2 into . To find the double angle formula, we use the Pythgorean identities.
Some double angle formula:
$\sin 2\theta = 2\sin \theta \cos \theta $
$\cos 2\theta = 1 - 2{\sin ^2}\theta = 2{\cos ^2}\theta - 1 = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
$\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$

Note: This type of question is usually solved by using the double angle formula. Here we can convert the expression into tan function. Then our aim is to apply a suitable double-angle formula. In this question, we will apply $\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$. Then substitute the value of $\cos {30^ \circ }$ to get the value of the given expression.