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 Question:
The question asks to simplify an expression involving inverse tangent functions and solve for a variable \( \alpha \).
Step 2: Key Formula or Approach:
1. \( \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2} \) for \( x>0 \).
2. \( \tan^{-1}(1/x) = \cot^{-1}(x) \) for \( x>0 \).
3. \( \tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \).
Step 3: Detailed Explanation:
LHS \( = \tan^{-1}(1) + [\tan^{-1}(4) + \tan^{-1}(1/4)] + \tan^{-1}(5) \)
Using \( \tan^{-1}(1/4) = \cot^{-1}(4) \):
LHS \( = \frac{\pi}{4} + \frac{\pi}{2} + \tan^{-1}(5) \)
LHS \( = \frac{3\pi}{4} + \tan^{-1}(5) \)
We need the RHS in the form \( \pi + \dots \).
LHS \( = \pi - \frac{\pi}{4} + \tan^{-1}(5) \)
LHS \( = \pi + \tan^{-1}(5) - \tan^{-1}(1) \)
Apply formula:
LHS \( = \pi + \tan^{-1}\left(\frac{5-1}{1+5 \cdot 1}\right) = \pi + \tan^{-1}\left(\frac{4}{6}\right) = \pi + \tan^{-1}\left(\frac{2}{3}\right) \)
Comparing with \( \pi + \tan^{-1}(\alpha/2) \):
\( \alpha/2 = 2/3 \Rightarrow \alpha = 4/3 \).
Step 4: Final Answer:
The value of \( \alpha \) is \( 4/3 \).