The quadratic equation, whose roots are ${\sin ^2}{18^ \circ }$and ${\cos ^2}{36^ \circ }$, is
A. $16{x^2} - 12x + 1 = 0$
B. $16{x^2} + 12x + 1 = 0$
C. $16{x^2} - 12x - 1 = 0$
D. $16{x^2} + 10x + 1 = 0$

Hint: In the given question, to find the quadratic equation whose roots are given i.e., ${\sin ^2}{18^ \circ }$ and ${\cos ^2}{36^ \circ }$. Find the sum and product of the roots for this put the value of $\sin {18^ \circ }$ and $\cos {36^ \circ }$. Lastly put the required values in ${x^2} - Px + Q = 0$, at the place of $P$ write the value of sum and for $Q$ write the value of product.

Formula Used:
Quadratic equation –
${x^2} - Px + Q = 0$

Complete step by step solution:
Given that,
The roots of quadratic equation are ${\sin ^2}{18^ \circ }$ and ${\cos ^2}{36^ \circ }$
Quadratic equation will be,
${x^2} - Px + Q = 0$
Where $P$ and $Q$ are the sum and product of the roots respectively.
${x^2} - \left[ {{{\sin }^2}{{18}^ \circ } + {{\cos }^2}{{36}^ \circ }} \right]x + \left[ {{{\sin }^2}{{18}^ \circ } \times {{\cos }^2}{{36}^ \circ }} \right] = 0$
${x^2} - \left[ {{{\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)}^2} + {{\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)}^2}} \right]x + \left[ {{{\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)}^2} \times {{\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)}^2}} \right] = 0$
${x^2} - \left[ {\dfrac{1}{{16}}{{\left( {\sqrt 5 - 1} \right)}^2} + {{\left( {\sqrt 5 + 1} \right)}^2}} \right]x + {\left[ {\dfrac{{5 - 4}}{{16}}} \right]^2} = 0$
${x^2} - \left[ {\dfrac{3}{4}} \right]x + \left[ {\dfrac{1}{{16}}} \right] = 0$
$16{x^2} - 12x + 1 = 0$

Option ‘A’ is correct

Note: The key concept involved in solving this problem is the good knowledge of quadratic equations. Students must know that if roots are given then we can directly find the equation using ${x^2} - Px + Q = 0$where $P$ and $Q$ are the sum and product of the roots respectively. Likewise, if the equation is given as $a{x^2} + bx + c = 0$and we have to find the sum and product of roots we can find directly using Sum of roots $ = \dfrac{{ - b}}{a}$ and product $ = \dfrac{c}{a}$.