Trigonometrical Ratios and Identities 1 Question 20
20. If $\frac{\sin ^{4} x}{2}+\frac{\cos ^{4} x}{3}=\frac{1}{5}$, then
(2009)
(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) $\frac{\sin ^{8} x}{8}+\frac{\cos ^{8} x}{27}=\frac{2}{125}$
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Solution:
- $\frac{\sin ^{4} x}{2}+\frac{\cos ^{4} x}{3}=\frac{1}{5} \Rightarrow \frac{\sin ^{4} x}{2}+\frac{\left(1-\sin ^{2} x\right)^{2}}{3}=\frac{1}{5}$
$\Rightarrow \quad \frac{\sin ^{4} x}{2}+\frac{1+\sin ^{4} x-2 \sin ^{2} x}{3}=\frac{1}{5}$
$\Rightarrow \quad 5 \sin ^{4} x-4 \sin ^{2} x+2=\frac{6}{5}$
$\Rightarrow \quad 25 \sin ^{4} x-20 \sin ^{2} x+4=0$
$$ \Rightarrow \quad\left(5 \sin ^{2} x-2\right)^{2}=0 $$
$$ \begin{array}{rlrl} \Rightarrow & \sin ^{2} x & =\frac{2}{5} \\ \cos ^{2} x & =\frac{3}{5}, \tan ^{2} x=\frac{2}{3} \\ \therefore \quad & \frac{\sin ^{8} x}{8}+\frac{\cos ^{8} x}{27} & =\frac{1}{125} \end{array} $$
