Question

The value of ${\cot ^2}{36^ \circ }{\cot ^2}{72^ \circ }$ is:
(A) $\dfrac{1}{2}$
(B) $\dfrac{1}{3}$
(C) $\dfrac{1}{4}$
(D) $\dfrac{1}{5}$

Answer 1

Hint : The given question deals with basic simplification of trigonometric e by using some of the simple trigonometric formulae such as $\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}}$ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem.

Complete step-by-step answer :
In the given problem, we have to simplify the product ${\cot ^2}{36^ \circ }{\cot ^2}{72^ \circ }$ .
So, ${\cot ^2}{36^ \circ }{\cot ^2}{72^ \circ }$
We first express the entire trigonometric expression in the terms of sine and cosine so as to simplify the expression.
We know the trigonometric formula $\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}}$ . So, we get,
$\Rightarrow \dfrac{{{{\cos }^2}{{36}^ \circ }}}{{{{\sin }^2}{{36}^ \circ }}} \times \dfrac{{{{\cos }^2}{{72}^ \circ }}}{{{{\sin }^2}{{72}^ \circ }}}$
Now, we know the values of the trigonometric function for the angles ${36^ \circ }$ and ${72^ \circ }$ . Hence, we can substitute the values and simplify the expression further.
So, putting in the value of $\cos \left( {{{36}^ \circ }} \right)$ as $\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)$ , value of $\sin \left( {{{36}^ \circ }} \right)$ as $\left( {\dfrac{{\sqrt {10 - 2\sqrt 5 } }}{4}} \right)$ , value of $\sin \left( {{{72}^ \circ }} \right)$ as $\left( {\dfrac{{\sqrt {10 + 2\sqrt 5 } }}{4}} \right)$ and the value of $\cos \left( {{{72}^ \circ }} \right)$ as $\left( {\dfrac{{\sqrt {10 + 2\sqrt 5 } }}{4}} \right)$ . Hence, we get,
$\Rightarrow \dfrac{{{{\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)}^2}}}{{{{\left( {\dfrac{{\sqrt {10 - 2\sqrt 5 } }}{4}} \right)}^2}}} \times \dfrac{{{{\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)}^2}}}{{{{\left( {\dfrac{{\sqrt {10 + 2\sqrt 5 } }}{4}} \right)}^2}}}$
Now, we have to evaluate the squares of the terms required and simplify the expression.
$\Rightarrow \dfrac{{\left( {\dfrac{{5 + 2\sqrt 5 + 1}}{{16}}} \right)}}{{\left( {\dfrac{{10 - 2\sqrt 5 }}{{16}}} \right)}} \times \dfrac{{\left( {\dfrac{{5 - 2\sqrt 5 + 1}}{{16}}} \right)}}{{\left( {\dfrac{{10 + 2\sqrt 5 }}{{16}}} \right)}}$
Cancelling the common factors and simplifying the expression further, we get,
$\Rightarrow \dfrac{{\left( {5 + 2\sqrt 5 + 1} \right)}}{{\left( {10 - 2\sqrt 5 } \right)}} \times \dfrac{{\left( {5 - 2\sqrt 5 + 1} \right)}}{{\left( {10 + 2\sqrt 5 } \right)}}$
$\Rightarrow \dfrac{{\left( {6 + 2\sqrt 5 } \right)}}{{\left( {10 - 2\sqrt 5 } \right)}} \times \dfrac{{\left( {6 - 2\sqrt 5 } \right)}}{{\left( {10 + 2\sqrt 5 } \right)}}$
Now, using the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ in the numerator and denominator, we get,
$\Rightarrow \dfrac{{\left( {6 + 2\sqrt 5 } \right)\left( {6 - 2\sqrt 5 } \right)}}{{\left( {10 - 2\sqrt 5 } \right)\left( {10 + 2\sqrt 5 } \right)}} = \dfrac{{{{\left( 6 \right)}^2} - {{\left( {2\sqrt 5 } \right)}^2}}}{{{{\left( {10} \right)}^2} - {{\left( {2\sqrt 5 } \right)}^2}}}$
$\Rightarrow \dfrac{{36 - 20}}{{100 - 20}}$
Simplifying the expression further, we get,
$\Rightarrow \dfrac{{16}}{{80}}$
$\Rightarrow \dfrac{1}{5}$
Hence, the value of the product ${\cot ^2}{36^ \circ }{\cot ^2}{72^ \circ }$ is $\left( {\dfrac{1}{5}} \right)$ by the use of basic algebraic rules and simple trigonometric formulae.
So, the correct answer is “Option D”.

Note : Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: $\tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}}$ and $\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}}$ . Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. One must now the values of trigonometric functions for the angles ${36^ \circ }$ and ${72^ \circ }$ in order to solve the problem.