Q31 / 60 · JEE Advanced 2009 (Paper 1) · Mathematics

If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$, then

  • Correct answer
    A
    $\tan ^2 x=\frac{2}{3}$
  • Correct answer
    B
    $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
  • C
    $\tan ^2 x=\frac{1}{3}$
  • D
    $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$

Solution

$$ \begin{aligned} & \frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5} \\ & \Rightarrow \quad \frac{\sin ^4 x}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5} \\ & \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5} \\ & \Rightarrow \quad 5 \sin ^4 x-4 \sin ^2 x+2=\frac{6}{5} \\ & \Rightarrow \quad 25 \sin ^4 x-20 \sin ^2 x+4=0 \\ & \Rightarrow \quad\left(5 \sin ^2 x-2\right)^2=0 \\ & \Rightarrow \sin ^2 x=\frac{2}{5}, \cos ^2 x=\frac{3}{5}, \tan ^2 x=\frac{2}{3} \\ & \therefore \quad \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125} \\ & \end{aligned} $$

Hence, the correct answer is options: 'A', 'B'.

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