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

\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}$$