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: Taking tangent of the RHS:
Let \(\theta=\tan^{-1}\dfrac{2}{11}+\tan^{-1}\dfrac{7}{24}\); we show \(\tan\theta=\dfrac12\).

Step 2: Applying tan(A+B):
\(\tan\theta=\dfrac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}\cdot\frac{7}{24}}=\dfrac{125/264}{250/264}=\dfrac12\).

Step 3: Checking the quadrant:
Both \(\tan^{-1}\frac{2}{11}\) and \(\tan^{-1}\frac{7}{24}\) are small positive acute angles, so their sum \(\theta\) is also acute; hence \(\theta=\tan^{-1}\frac12\) unambiguously.

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