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Question: If \(\dfrac{{{{\sin }^4}x}}{2} + \dfrac{{{{\cos }^4}x}}{3} = \dfrac{1}{5}\) ,then \( {\text{A...

If sin⁡4x2+cos⁡4x3=15\dfrac{{{{\sin }^4}x}}{2} + \dfrac{{{{\cos }^4}x}}{3} = \dfrac{1}{5} ,then
A. tan2x=43 B. sin⁡8x8+cos⁡8x27=1125 C. tan2x=13 D. sin⁡8x8+cos⁡8x27=2125  {\text{A}}{\text{. ta}}{{\text{n}}^2}x = \dfrac{4}{3} \\\ {\text{B}}{\text{. }}\dfrac{{{{\sin }^8}x}}{8} + \dfrac{{{{\cos }^8}x}}{{27}} = \dfrac{1}{{125}} \\\ {\text{C}}{\text{. ta}}{{\text{n}}^2}x = \dfrac{1}{3} \\\ {\text{D}}{\text{. }}\dfrac{{{{\sin }^8}x}}{8} + \dfrac{{{{\cos }^8}x}}{{27}} = \dfrac{2}{{125}} \\\

Explanation

Solution

As given in the question we have equation in sine and cosine function and we have to find tan so first we will simplify it using simple trigonometric calculations and find the value of individual sine and cosine function and then we can easily check all the options by putting individual values of sine and cosine function.

Complete step-by-step answer :
We have given
sin⁡4x2+cos⁡4x3=15\dfrac{{{{\sin }^4}x}}{2} + \dfrac{{{{\cos }^4}x}}{3} = \dfrac{1}{5}
For simple calculation first multiply whole equation by 6 then we get,
3sin⁡4x+2cos⁡4x=653{\sin ^4}x + 2{\cos ^4}x = \dfrac{6}{5}
And we know standard formula (cos⁡2x=1−sin⁡2x)\left( {{{\cos }^2}x = 1 - {{\sin }^2}x} \right)
Using this formula we can write above equation as
3sin⁡4x+2(1−sin⁡2x)2=653{\sin ^4}x + 2{\left( {1 - {{\sin }^2}x} \right)^2} = \dfrac{6}{5}
On further simplification we get
3sin⁡4x+2(1+sin⁡4x−2sin⁡2x)=653{\sin ^4}x + 2\left( {1 + {{\sin }^4}x - 2{{\sin }^2}x} \right) = \dfrac{6}{5}
On cross multiply we get,
15sin⁡4x+10(1+sin⁡4x−2sin⁡2x)=615{\sin ^4}x + 10\left( {1 + {{\sin }^4}x - 2{{\sin }^2}x} \right) = 6
On further simplifying we get,
15sin⁡4x+10+10sin⁡4x−20sin⁡2x=615{\sin ^4}x + 10 + 10{\sin ^4}x - 20{\sin ^2}x = 6
25sin⁡4x−20sin⁡2x+4=025{\sin ^4}x - 20{\sin ^2}x + 4 = 0
Now we will make in factor form
25sin⁡4x−10sin⁡2x−10sin⁡2x+4=0 5sin⁡2x(5sin⁡2x−2)−2(5sin⁡2x−2)=0 (5sin⁡2x−2)(5sin⁡2x−2)=0  25{\sin ^4}x - 10{\sin ^2}x - 10{\sin ^2}x + 4 = 0 \\\ 5{\sin ^2}x\left( {5{{\sin }^2}x - 2} \right) - 2\left( {5{{\sin }^2}x - 2} \right) = 0 \\\ \left( {5{{\sin }^2}x - 2} \right)\left( {5{{\sin }^2}x - 2} \right) = 0 \\\
And hence we get
sin⁡2x=25{\sin ^2}x = \dfrac{2}{5}
And we know
cos⁡2x=1−sin⁡2x ∴cos⁡2x=1−25=35  {\cos ^2}x = 1 - {\sin ^2}x \\\ \therefore {\cos ^2}x = 1 - \dfrac{2}{5} = \dfrac{3}{5} \\\

And hence
tan⁡2x=sin⁡2xcos⁡2x=2535=23{\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{{\dfrac{2}{5}}}{{\dfrac{3}{5}}} = \dfrac{2}{3}
But neither option A nor option C is correct so we will further check for option B and D.
We have in option B
sin⁡8x8+cos⁡8x27=1125\dfrac{{{{\sin }^8}x}}{8} + \dfrac{{{{\cos }^8}x}}{{27}} = \dfrac{1}{{125}}
We can write it as
(sin⁡2x)48+(cos⁡2x)427=1125\dfrac{{{{\left( {{{\sin }^2}x} \right)}^4}}}{8} + \dfrac{{{{\left( {{{\cos }^2}x} \right)}^4}}}{{27}} = \dfrac{1}{{125}}
Now on putting value of sin⁡2x,cos⁡2x{\sin ^2}x,{\cos ^2}x we get,
(25)48+(35)427=1125\dfrac{{{{\left( {\dfrac{2}{5}} \right)}^4}}}{8} + \dfrac{{{{\left( {\dfrac{3}{5}} \right)}^4}}}{{27}} = \dfrac{1}{{125}}
On further simplifying we get
16625×8+81625×27=1125\dfrac{{16}}{{625 \times 8}} + \dfrac{{81}}{{625 \times 27}} = \dfrac{1}{{125}}
2625+3625=1125 5625=1125 ∴1125=1125  \dfrac{2}{{625}} + \dfrac{3}{{625}} = \dfrac{1}{{125}} \\\ \dfrac{5}{{625}} = \dfrac{1}{{125}} \\\ \therefore \dfrac{1}{{125}} = \dfrac{1}{{125}} \\\
Hence option B is the correct option.

Note : Whenever we get this type of question the key concept of solving it is we have to remember all the trigonometric formulas and we have to learn to apply these formulas so that we can answer this type of question easily.