Question

Evaluate \[ \tan^{-1} \frac{\sqrt{8} - 2\sqrt{15}}{\sqrt{15} + 1} + \tan^{-1} \frac{1}{\sqrt{5}} \]

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{2}\)

Hint:

Use sum formula for inverse tangent to combine terms.

Answer 1

The Correct Option is A

Step 1: Use formula for sum of inverse tangents
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \] Step 2: Let \[ a = \frac{\sqrt{8} - 2\sqrt{15}}{\sqrt{15} + 1}, \quad b = \frac{1}{\sqrt{5}} \] Step 3: Calculate numerator
\[ a + b = \frac{\sqrt{8} - 2\sqrt{15}}{\sqrt{15} + 1} + \frac{1}{\sqrt{5}} \] Step 4: Calculate denominator
\[ 1 - ab = 1 - \left(\frac{\sqrt{8} - 2\sqrt{15}}{\sqrt{15} + 1}\right) \times \frac{1}{\sqrt{5}} \] Step 5: Simplify fraction
After simplification, the value equals \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\). Therefore, \[ \tan^{-1} a + \tan^{-1} b = \frac{\pi}{6} \]