First slide

Trigonometric Identities

Question

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

Moderate

A

$\tan^{2} x + \frac{4}{3}$
B

$\frac{\sin^{8} x}{8} + \frac{\cos^{8} x}{27} = \frac{1}{125}$
C

$\tan^{2} x = \frac{1}{3}$
D

$\frac{\sin^{8} x}{8} + \frac{\cos^{8} x}{27} = \frac{2}{125}$

Solution

$\begin{matrix} we have, \frac{\sin^{4} x}{2} + \frac{\cos^{4} x}{3} = \frac{1}{5} \\ \frac{\sin^{4} x}{2} + \frac{{(1 - \sin^{2} x)}^{2}}{3} = \frac{1}{5} \\ 3 \sin^{4} x + 2 {(1 - \sin^{2} x)}^{2} = \frac{6}{5} \\ 25 \sin^{4} x - 20 \sin^{2} x + 4 = 0 \\ \sin^{2} x = \frac{2}{5} \\ \cos^{2} x = 3 / 5 \\ \tan^{2} x = 2 / 3 \\ \frac{\sin^{8} x}{8} + \frac{\cos^{8} x}{27} = \frac{(2 / 5)^{4}}{8} + \frac{(3 / 5)^{4}}{27} \\ = \frac{2}{625} + \frac{3}{625} \\ = \frac{5}{625} = \frac{1}{125} \end{matrix}$

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