Question

Evaluate: \(\tan^{-1}(1) + \tan^{-1}(4) + \tan^{-1}(5) + \tan^{-1}\left(\frac{1}{4}\right) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right)\). Find the value of \(\alpha\).

Hint:

Always look for reciprocal pairs \((x, 1/x)\) in \(\tan^{-1}\) sums. Their sum is always \(\pi/2\). This is a favorite trick in entrance exams to test your observation skills.

Answer 1

Step 1: Understanding the Concept:
We use the property that \(\tan^{-1}(x) + \tan^{-1}(1/x) = \pi/2\) for \(x>0\). This simplifies the series significantly.

Step 2: Key Formula or Approach:
1. \(\tan^{-1}(1) = \pi/4\).
2. \(\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}(\frac{x+y}{1-xy})\) (used here for subtraction).

Step 3: Detailed Explanation:
Let the LHS be \(L\):
\[ L = \tan^{-1}(1) + \tan^{-1}(4) + \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}(5) \] Using \(\tan^{-1}(4) + \tan^{-1}(1/4) = \pi/2\):
\[ L = \frac{\pi}{4} + \frac{\pi}{2} + \tan^{-1}(5) = \frac{3\pi}{4} + \tan^{-1}(5) \] The equation is \(L = \pi + \tan^{-1}(\alpha/2)\).
\[ \frac{3\pi}{4} + \tan^{-1}(5) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right) \] \[ \tan^{-1}\left(\frac{\alpha}{2}\right) = \tan^{-1}(5) - \frac{\pi}{4} \] Since \(\pi/4 = \tan^{-1}(1)\):
\[ \tan^{-1}\left(\frac{\alpha}{2}\right) = \tan^{-1}(5) - \tan^{-1}(1) \] Using the formula \(\tan^{-1}x - \tan^{-1}y = \tan^{-1}(\frac{x-y}{1+xy})\):
\[ \tan^{-1}\left(\frac{\alpha}{2}\right) = \tan^{-1}\left(\frac{5 - 1}{1 + 5(1)}\right) = \tan^{-1}\left(\frac{4}{6}\right) = \tan^{-1}\left(\frac{2}{3}\right) \] By comparison:
\[ \frac{\alpha}{2} = \frac{2}{3} \implies \alpha = \frac{4}{3} \]
Step 4: Final Answer:
The value of \(\alpha\) is \(\frac{4}{3}\).