A) 1/3 and 3
B) 1/4 and 4
C) 1/5 and 5
D) 5 and 6
Correct Answer: A
Solution :
Let \[y=\frac{\tan x}{\tan 3x}=\frac{\tan x}{\frac{3\tan x-{{\tan }^{3}}x}{1-3{{\tan }^{2}}x}}\] \[y=\frac{1-3{{\tan }^{2}}x}{3-{{\tan }^{2}}x}=\frac{\frac{1}{3}-{{\tan }^{2}}x}{1-\frac{1}{3}.{{\tan }^{2}}x}\] Hence, y should never lie between \[\frac{1}{3}\]and 3 whenever defined.
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