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

Hint: Change the whole equation as in either of \[\sin x\] or \[\cos x\] form then solve the equation as an quadratic equation and get either of \[\sin x\] or \[\cos x\] and finally divide them to get the value of \[\tan x\] after all this examine the options carefully.

Complete step-by-step answer:
Let us start by taking LCM we will get it as
\[\begin{array}{l}
\therefore \dfrac{{{{\sin }^4}x}}{2} + \dfrac{{{{\cos }^4}x}}{3} = \dfrac{1}{5}\\
 \Rightarrow \dfrac{{3{{\sin }^4}x + 2{{\cos }^4}x}}{6} = \dfrac{1}{5}\\
 \Rightarrow 3{\sin ^4}x + 2{\cos ^4}x = \dfrac{6}{5}
\end{array}\]
Now we know that \[{{{\sin }^2}x + {{\cos }^2}x = 1}\]
Using this we will get it as
\[\begin{array}{l}
 \Rightarrow 3{\sin ^4}x + 2{\left( {1 - {{\sin }^2}x} \right)^2} = \dfrac{6}{5}\\
 \Rightarrow 3{\sin ^4}x + 2\left( {1 + {{\sin }^4}x - 2{{\sin }^2}x} \right) = \dfrac{6}{5}\\
 \Rightarrow 5{\sin ^4}x - 4{\sin ^2}x + 2 = \dfrac{6}{5}\\
 \Rightarrow 25{\sin ^4}x - 20{\sin ^2}x + 4 = 0\\
 \Rightarrow 25{\sin ^4}x - 10{\sin ^2}x - 10{\sin ^2}x + 4 = 0\\
 \Rightarrow 5{\sin ^2}x\left( {5{{\sin }^2}x - 2} \right) - 2\left( {5{{\sin }^2}x - 2} \right) = 0\\
 \Rightarrow {\left( {5{{\sin }^2}x - 2} \right)^2} = 0\\
\therefore 5{\sin ^2}x - 2 = 0\\
 \Rightarrow 5{\sin ^2}x = 2\\
 \Rightarrow {\sin ^2}x = \dfrac{2}{5}
\end{array}\]
Now as we have the value of \[{\sin ^2}x\]
\[\begin{array}{l}
\therefore {\cos ^2}x = 1 - {\sin ^2}x\\
 \Rightarrow {\cos ^2}x = 1 - \dfrac{2}{5}\\
 \Rightarrow {\cos ^2}x = \dfrac{3}{5}
\end{array}\]
Now as we have the value of \[{\cos ^2}x\& {\sin ^2}x\]
\[\begin{array}{l}
\therefore {\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\\
 \Rightarrow {\tan ^2}x = \dfrac{{\dfrac{2}{5}}}{{\dfrac{3}{5}}}\\
 \Rightarrow {\tan ^2}x = \dfrac{2}{3}
\end{array}\]
Clearly option A is correct here and option C is incorrect
Let us check the other option also
We know that \[{\left( {{y^2}} \right)^3} = {y^8}\]
So same goes for \[{\cos ^2}x\& {\sin ^2}x\]
\[\therefore \dfrac{{{{\sin }^2}x}}{2} = \dfrac{1}{5}\& \dfrac{{{{\cos }^2}x}}{3} = \dfrac{1}{5}\]
So if we just cube and add them we will get it as
\[\begin{array}{l}
 = \dfrac{{{{\sin }^8}x}}{8} + \dfrac{{{{\cos }^8}x}}{{27}}\\
 = {\left( {\dfrac{1}{5}} \right)^3} + {\left( {\dfrac{1}{5}} \right)^3}\\
 = \dfrac{1}{{125}} + \dfrac{1}{{125}}\\
 = \dfrac{2}{{125}}\\
\therefore \dfrac{{{{\sin }^8}x}}{8} + \dfrac{{{{\cos }^8}x}}{{27}} = \dfrac{2}{{125}}
\end{array}\]
So clearly here Option D is correct and option B is incorrect.
Which means options A and D both are correct.

Note: While solving \[{\left( {1 - {{\sin }^2}x} \right)^2}\] i have used the arithmetic formula \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] also note that \[\tan x = \dfrac{p}{b},\sin x = \dfrac{p}{h},\cos x = \dfrac{b}{h}\] thats why \[\tan x = \dfrac{{\sin x}}{{\cos x}} = \dfrac{{\dfrac{p}{h}}}{{\dfrac{b}{h}}} = \dfrac{p}{b}\]