JEE Advanced · Mathematics · 8. Trigonometric Equations
If \(\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}\), then
- A \(\tan ^2 x=\frac{2}{3}\)
- B \(\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}\)
- C \(\tan ^2 x=\frac{1}{3}\)
- D \(\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}\)
Answer & Solution
Correct Answer
(B) \(\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}\)
Step-by-step Solution
Detailed explanation
\(\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5} \)
\( \Rightarrow \frac{\sin ^4 x}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5} \)
\( \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5} \)
\( \Rightarrow 5 \sin ^4 x-4 \sin ^2 x+2=\frac{6}{5} \)
\( \Rightarrow 25 \sin ^4 x-20 \sin ^2 x+4=0 \)
\( \Rightarrow \left(5 \sin ^2 x-2\right)^2=0 \)
\( \Rightarrow \sin ^2 x=\frac{2}{5}, \cos ^2 x=\frac{3}{5}, \tan ^2 x=\frac{2}{3} \)
\( \therefore \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}\)
\( \Rightarrow \frac{\sin ^4 x}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5} \)
\( \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5} \)
\( \Rightarrow 5 \sin ^4 x-4 \sin ^2 x+2=\frac{6}{5} \)
\( \Rightarrow 25 \sin ^4 x-20 \sin ^2 x+4=0 \)
\( \Rightarrow \left(5 \sin ^2 x-2\right)^2=0 \)
\( \Rightarrow \sin ^2 x=\frac{2}{5}, \cos ^2 x=\frac{3}{5}, \tan ^2 x=\frac{2}{3} \)
\( \therefore \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}\)
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