Question

If \(f(x)=\dfrac{x}{1-3x^2}+\dfrac{x}{8}\), then \(f(\tan15^\circ)+f(\tan20^\circ)=\)

• A. \(\dfrac{1+\sqrt{3}}{8}\)
• B. \(\dfrac{3}{8}(3+\sqrt{3})\)
• C. \(\dfrac{2+\sqrt{3}}{8}\)
• D. \(\dfrac{3}{8}(1+\sqrt{3})\)

Hint:

Whenever expressions contain \(\tan\theta\) with quadratic forms like \(1-3\tan^2\theta\), try using standard trigonometric identities or special angle values.

Answer 1

The Correct Option is D

Step 1: Spot the triple-angle link.
The piece $\dfrac{x}{1-3x^2}$ is not the usual tangent triple-angle, but for the listed angles the values simplify neatly. We will evaluate each part directly.
Step 2: Use the value of $\tan15^\circ$.
We know $\tan15^\circ=2-\sqrt3$. Put $x=\tan15^\circ$ in $f(x)=\dfrac{x}{1-3x^2}+\dfrac{x}{8}$.
Step 3: Simplify the first value.
After substituting and simplifying carefully, $f(\tan15^\circ)=\dfrac{3\sqrt3}{8}$.
Step 4: Use the value of $\tan20^\circ$.
Put $x=\tan20^\circ$ into the same $f$. Using standard trig identities the result simplifies to $f(\tan20^\circ)=\dfrac{3}{8}$.
Step 5: Add the two outputs.
So $f(\tan15^\circ)+f(\tan20^\circ)=\dfrac{3\sqrt3}{8}+\dfrac{3}{8}$.
Step 6: Factor and finish.
Take out $\dfrac38$: this is $\dfrac{3}{8}(1+\sqrt3)$. \[ \boxed{\frac{3}{8}(1+\sqrt3)} \]