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: Another grouping
Pair $\tan^{-1}1$ with $\tan^{-1}\frac14$ and $\tan^{-1}3$ with $\tan^{-1}5$.
$\tan^{-1}1+\tan^{-1}\frac14=\tan^{-1}\frac{5/4}{3/4}=\tan^{-1}\frac53$ (product $<1$).
$\tan^{-1}3+\tan^{-1}5=\pi+\tan^{-1}\frac{8}{-14}=\pi-\tan^{-1}\frac47$.

Step 2: Add
$\tan^{-1}\frac53-\tan^{-1}\frac47=\tan^{-1}\frac{\frac53-\frac47}{1+\frac{20}{21}}=\tan^{-1}\frac{23/21}{41/21}=\tan^{-1}\frac{23}{41}$.
So the sum is $\pi+\tan^{-1}\frac{23}{41}$, giving $\alpha=\frac{46}{41}$.

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