Question

Show that \(\tan^{-1}\dfrac{2}{11}+\tan^{-1}\dfrac{7}{24}=\tan^{-1}\dfrac{1}{2}\).

Hint:

Use tan⁻¹a + tan⁻¹b = tan⁻¹((a+b)/(1−ab)) since ab < 1 here.

Answer 1

Step 1: Using the addition formula:
For \(ab<1\), \(\tan^{-1}a+\tan^{-1}b=\tan^{-1}\dfrac{a+b}{1-ab}\). Here \(a=\dfrac{2}{11}\), \(b=\dfrac{7}{24}\), and \(ab=\dfrac{14}{264}<1\), so the formula applies directly.

Step 2: Computing the sum and product:
\(a+b=\dfrac{2}{11}+\dfrac{7}{24}=\dfrac{48+77}{264}=\dfrac{125}{264}\), and \(1-ab=1-\dfrac{14}{264}=\dfrac{250}{264}\).

Step 3: Simplifying the ratio:
\(\dfrac{a+b}{1-ab}=\dfrac{125/264}{250/264}=\dfrac{125}{250}=\dfrac{1}{2}\).

Final Answer:
\(\tan^{-1}\dfrac{2}{11}+\tan^{-1}\dfrac{7}{24}=\tan^{-1}\dfrac{1}{2}\).\[ \boxed{\tan^{-1}\tfrac12} \]