Question

If $\dfrac{{{\sin }^{4}}x}{2}+\dfrac{{{\cos }^{4}}x}{3}=\dfrac{1}{5}$, then
(a)${{\tan }^{2}}x=\dfrac{2}{3}$
(b) $\dfrac{{{\sin }^{8}}x}{8}+\dfrac{{{\cos }^{8}}x}{27}=\dfrac{1}{125}$
(c) ${{\tan }^{2}}x=\dfrac{1}{3}$
(d) $\dfrac{{{\sin }^{8}}x}{8}+\dfrac{{{\cos }^{8}}x}{27}=\dfrac{2}{125}$

Answer 1

We convert the given equation into an equation with only one unknown, that is, substitute ${{\cos }^{4}}x$ with ${{(1-{{\sin }^{2}}x)}^{2}}$. Replace ${{\sin }^{2}}x$ by $a$to obtain a quadratic equation in $a$. Solve this equation to get the value of $a$ and thereby ${{\sin }^{2}}x$. Using different trigonometric relations, find the values of ${{\cos }^{2}}x$ and ${{\tan }^{2}}x$ to obtain the correct answer.
Use the following trigonometric identities:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
$\dfrac{\sin x}{\cos x}=\tan x$
$\dfrac{1}{\sec x}=\cos x$
${{\sec }^{2}}x=1+{{\tan }^{2}}x$
Find the values according to the question and check for the option satisfying the obtained value(s).

Complete step-by-step answer:
We have,
$\dfrac{{{\sin }^{4}}x}{2}+\dfrac{{{\cos }^{4}}x}{3}=\dfrac{1}{5}.................(1)$
We know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
On rearranging, we get,
${{\cos }^{2}}x=1-{{\sin }^{2}}x$
Substituting for ${{\cos }^{2}}x$ in equation (1), we obtain,
$\dfrac{{{\sin }^{4}}x}{2}+\dfrac{{{(1-{{\sin }^{2}}x)}^{2}}}{3}=\dfrac{1}{5}$
On taking LCM, we get,
$\dfrac{3{{\sin }^{4}}x}{6}+\dfrac{2{{(1-{{\sin }^{2}}x)}^{2}}}{6}=\dfrac{1}{5}$
On cross-multiplying,
$5\left( 3{{\sin }^{4}}x+2{{(1-{{\sin }^{2}}x)}^{2}} \right)=6$
$15{{\sin }^{4}}x+10{{(1-{{\sin }^{2}}x)}^{2}}=6$
For ease of simplification, let us replace ${{\sin }^{2}}x$with $a$ , which we can replace back later on.
$15{{a}^{2}}+10{{(1-a)}^{2}}=6$
Using the algebraic identity ${{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$ the equation becomes,
$15{{a}^{2}}+10(1-2a+{{a}^{2}})=6$
$15{{a}^{2}}+10-20a+10{{a}^{2}}=6$
$25{{a}^{2}}-20a+4=0$
We now have to solve the obtained quadratic equation.
We have three different methods to solve a quadratic equation namely factoring or splitting the middle term, completing the square method and the quadratic formula (Using discriminant). Here we are solving the quadratic equation using factoring method.
Using splitting the middle term method, as the name suggests we will split the middle term into two, whose sum is equal to the middle term and their product is equal to the product of the first and third term.
In the quadratic equation $25{{a}^{2}}-20a+4=0$, the product is equal to $100{{a}^{2}}$ and the sum is equal to $-20a$.
Hence, we split the equation as,
$25{{a}^{2}}-10a-10a+4=0$
$5a(5a-2)-2(5a-2)=0$
${{(5a-2)}^{2}}=0$
$5a-2=0$
$\therefore a=\dfrac{2}{5}$
But $a={{\sin }^{2}}x$.
$\therefore {{\sin }^{2}}x=\dfrac{2}{5}............(2)$
We know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
On rearranging, we get,
${{\cos }^{2}}x=1-{{\sin }^{2}}x$
$\therefore {{\cos }^{2}}x=1-\dfrac{2}{5}=\dfrac{3}{5}..................(3)$
On dividing equation (2) by equation (3), we get,
${{\tan }^{2}}x=\dfrac{\left( \dfrac{2}{5} \right)}{\left( \dfrac{3}{5} \right)}$
$\therefore {{\tan }^{2}}x=\dfrac{2}{3}$

So, the correct answer is “Option A”.

Note: Alternate method:
The equation $\dfrac{{{\sin }^{4}}x}{2}+\dfrac{{{\cos }^{4}}x}{3}=\dfrac{1}{5}$ can be solved in another method.
Dividing through by ${{\cos }^{4}}x$, the equation becomes,
$\dfrac{{{\sin }^{4}}x}{2{{\cos }^{4}}x}+\dfrac{1}{3}=\dfrac{1}{5\left( {{\cos }^{4}}x \right)}$
We know that, $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{1}{\sec x}=\cos x$
Hence, the equation becomes,
$\dfrac{{{\tan }^{4}}x}{2}+\dfrac{1}{3}=\dfrac{{{\sec }^{4}}x}{5}$
Substituting ${{\sec }^{2}}x=1+{{\tan }^{2}}x$ in the above equation, it becomes,
$\dfrac{{{\tan }^{4}}x}{2}+\dfrac{1}{3}=\dfrac{{{(1+{{\tan }^{2}}x)}^{2}}}{5}$
On replacing ${{\tan }^{2}}x$ by a, and solving the obtained quadratic equation, we get the value of $a$ and hence ${{\tan }^{2}}x$.