Question

If \(tan^{-1}(1)+tan^{-1}(3)+tan^{-1}(5)+tan^{-1}(\frac{1}{4}) = π+tan^{-1}(\frac{α}{2})\), then the value of \(α\) is...

• A. \(\frac{46}{41}\)
• B. \(\frac{23}{41}\)
• C. \(\frac{42}{41}\)
• D. \(\frac{44}{41}\)

Hint:

Combine the inverse tangents two at a time and watch the $\pi$ corrections.

Answer 1

The Correct Option is A

Step 1: Combine \(\tan^{-1}1+\tan^{-1}3\)
Since \(1\times3>1\), the sum is \(\pi+\tan^{-1}\frac{1+3}{1-3}=\pi+\tan^{-1}(-2)=\pi-\tan^{-1}2\).

Step 2: Combine \(\tan^{-1}5+\tan^{-1}\frac14\)
\(5\times\frac14=\frac54>1\), so the sum is \(\pi+\tan^{-1}\frac{5+\frac14}{1-\frac54}=\pi+\tan^{-1}(-21)=\pi-\tan^{-1}21\).

Step 3: Add them
Total \(=2\pi-(\tan^{-1}2+\tan^{-1}21)\). Here \(2\times21>1\), so \(\tan^{-1}2+\tan^{-1}21=\pi+\tan^{-1}\frac{23}{1-42}=\pi-\tan^{-1}\frac{23}{41}\).

Step 4: Result
Total \(=2\pi-\pi+\tan^{-1}\frac{23}{41}=\pi+\tan^{-1}\frac{23}{41}\).
Compare with \(\pi+\tan^{-1}\frac{\alpha}{2}\): \(\frac{\alpha}{2}=\frac{23}{41}\), so \(\alpha=\frac{46}{41}\).

Final Answer:
\(\alpha=\frac{46}{41}\), option (A). \[ \boxed{\text{(A) } \frac{46}{41}} \]