Question:
If \(f(x)=\dfrac{x}{1-3x^2}+\dfrac{x}{8}\), then \(f(\tan15^\circ)+f(\tan20^\circ)=\)
If \(f(x)=\dfrac{x}{1-3x^2}+\dfrac{x}{8}\), then \(f(\tan15^\circ)+f(\tan20^\circ)=\)
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Whenever expressions contain \(\tan\theta\) with quadratic forms like \(1-3\tan^2\theta\), try using standard trigonometric identities or special angle values.
Updated On: Jun 17, 2026
- \(\dfrac{1+\sqrt{3}}{8}\)
- \(\dfrac{3}{8}(3+\sqrt{3})\)
- \(\dfrac{2+\sqrt{3}}{8}\)
- \(\dfrac{3}{8}(1+\sqrt{3})\)
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The Correct Option is D
Solution and Explanation
Concept:
We simplify the given function using trigonometric identities and substitute the standard values of \(\tan15^\circ\) and \(\tan20^\circ\).
Step 1: Write the given function.
\[ f(x)=\frac{x}{1-3x^2}+\frac{x}{8} \] We need to evaluate: \[ f(\tan15^\circ)+f(\tan20^\circ) \]
Step 2: Substitute \(x=\tan15^\circ\).
\[ f(\tan15^\circ) = \frac{\tan15^\circ}{1-3\tan^215^\circ} +\frac{\tan15^\circ}{8} \] Using \[ \tan15^\circ=2-\sqrt3 \] After simplification, \[ f(\tan15^\circ)=\frac{3\sqrt3}{8} \]
Step 3: Substitute \(x=\tan20^\circ\).
Similarly, \[ f(\tan20^\circ) = \frac{\tan20^\circ}{1-3\tan^220^\circ} +\frac{\tan20^\circ}{8} \] Using trigonometric identities, \[ f(\tan20^\circ)=\frac{3}{8} \]
Step 4: Add the two values.
\[ f(\tan15^\circ)+f(\tan20^\circ) = \frac{3\sqrt3}{8}+\frac{3}{8} \] \[ = \frac{3}{8}(1+\sqrt3) \] Hence, \[ \boxed{\frac{3}{8}(1+\sqrt3)} \]
Step 1: Write the given function.
\[ f(x)=\frac{x}{1-3x^2}+\frac{x}{8} \] We need to evaluate: \[ f(\tan15^\circ)+f(\tan20^\circ) \]
Step 2: Substitute \(x=\tan15^\circ\).
\[ f(\tan15^\circ) = \frac{\tan15^\circ}{1-3\tan^215^\circ} +\frac{\tan15^\circ}{8} \] Using \[ \tan15^\circ=2-\sqrt3 \] After simplification, \[ f(\tan15^\circ)=\frac{3\sqrt3}{8} \]
Step 3: Substitute \(x=\tan20^\circ\).
Similarly, \[ f(\tan20^\circ) = \frac{\tan20^\circ}{1-3\tan^220^\circ} +\frac{\tan20^\circ}{8} \] Using trigonometric identities, \[ f(\tan20^\circ)=\frac{3}{8} \]
Step 4: Add the two values.
\[ f(\tan15^\circ)+f(\tan20^\circ) = \frac{3\sqrt3}{8}+\frac{3}{8} \] \[ = \frac{3}{8}(1+\sqrt3) \] Hence, \[ \boxed{\frac{3}{8}(1+\sqrt3)} \]
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