Question

The value of $\frac{tanx}{tan3x}$ whenever defined never lie between

• A. $\frac{1}{3}and3$
• B. $\frac{1}{4}and4$
• C. $\frac{1}{5}and5$
• D. $5and6$

Answer 1

The correct option is A

$\frac{1}{3}and3$

$Lety=\frac{tanx}{tan3x}=\frac{tanx}{\frac{3tanx-ta{n}^{3}x}{1-3ta{n}^{3}x}}y=\frac{1-3ta{n}^{2}x}{3-ta{n}^{2}x}=\frac{\frac{1}{3}-ta{n}^{2}x}{1-\frac{1}{3}.ta{n}^{2}x}$

Hence, y should never lie between $\frac{1}{3}$ and 3 whenever defined.