Question

If ${{\sin }^{4}}x+{{\sin }^{2}}x=1$ , then value of ${{\cos }^{4}}x+{{\cos }^{2}}x$ is
(A) $5-2\sqrt{5}$
(B)${{\sin }^{2}}x$
(C)${{\tan }^{2}}x$
(D)$1$

Answer 1

For solving this question we will use the formula of finding roots of quadratic equation $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and find the value of ${{\sin }^{2}}x$ from the equation we have ${{\sin }^{4}}x+{{\sin }^{2}}x=1$ and use it find the value of ${{\cos }^{2}}x$ since ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ and use this value in the equation ${{\cos }^{4}}x+{{\cos }^{2}}x$ and get the answer.

Complete step by step answer:
From the question we have the ${{\sin }^{4}}x+{{\sin }^{2}}x=1$. By transferring $1$ from the R.H.S to L.H.S we will have ${{\sin }^{4}}x+{{\sin }^{2}}x-1=0$ , a quadratic equation.
Since we know that for a quadratic equation $a{{x}^{2}}+bx+c=0$ the roots are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . By using this we can extract the value of ${{\sin }^{2}}x$ . The value of ${{\sin }^{2}}x$ is given as $\dfrac{-1\pm \sqrt{1-4\left( -1 \right)}}{2}$ . After simplifying this we will have $\dfrac{-1\pm \sqrt{1+4}}{2}\Rightarrow \dfrac{-1\pm \sqrt{5}}{2}$ .
Since squares of any value can’t be negative so we will have ${{\sin }^{2}}x=\dfrac{-1+\sqrt{5}}{2}$ .
Since we know that ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ we will have ${{\cos }^{2}}x=1-\dfrac{-1+\sqrt{5}}{2}$.
By performing the simplifications we will have ${{\cos }^{2}}x=\dfrac{3-\sqrt{5}}{2}$ .
By substituting this value in the equation we need to derive that is ${{\cos }^{4}}x+{{\cos }^{2}}x$.
After substituting we will have ${{\left( \dfrac{3-\sqrt{5}}{2} \right)}^{2}}+\dfrac{3-\sqrt{5}}{2}$ .
After expanding this we will have $\left( \dfrac{9-6\sqrt{5}+5}{4} \right)+\dfrac{3-\sqrt{5}}{2}$ .
After simplifying we will have
$\begin{aligned} & \left( \dfrac{14-6\sqrt{5}}{4} \right)+\dfrac{3-\sqrt{5}}{2}=\left( \dfrac{7-3\sqrt{5}}{2} \right)+\dfrac{3-\sqrt{5}}{2} \\\ & \Rightarrow \dfrac{10-4\sqrt{5}}{2}\Rightarrow 5-2\sqrt{5} \\\ \end{aligned}$.
Now we have a conclusion that when ${{\sin }^{4}}x+{{\sin }^{2}}x=1$ then ${{\cos }^{4}}x+{{\cos }^{2}}x=5-2\sqrt{5}$ .

So, the correct answer is “Option A”.

Note: While answering questions of this type we should remember that the square of any value can’t be negative and $\sqrt{5}=2.23$ so we can’t use the value of ${{\sin }^{2}}x$ as $\dfrac{-1-\sqrt{5}}{2}$ because it is a negative number. If we do that we will have a wrong answer.