Trigonometrical Ratios And Identities Ques 23
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) $\frac{\sin ^{8} x}{8}+\frac{\cos ^{8} x}{27}=\frac{2}{125}$
Correct Answer: 23.(a,b) Solution: $\Rightarrow \quad \frac{\sin ^{4} x}{2}+\frac{1+\sin ^{4} x-2 \sin ^{2} x}{3}=\frac{1}{5}$ $\Rightarrow \quad 5 \sin ^{4} x-4 \sin ^{2} x+2=\frac{6}{5}$ $\Rightarrow \quad 25 \sin ^{4} x-20 \sin ^{2} x+4=0$ $
\Rightarrow \quad \left(5 \sin ^{2} x-2\right)^{2}=0
$ $\Rightarrow \quad \sin ^{2} x =\frac{2}{5} $ $\cos ^{2} x =\frac{3}{5}, \tan ^{2} x=\frac{2}{3} $ $\therefore \quad \frac{\sin ^{8} x}{8}+\frac{\cos ^{8} x}{27} =\frac{1}{125}$Show Answer
Answer:
Formula:
BETA
