Question

Prove that \(\tan^{-1}\dfrac15+\tan^{-1}\dfrac17+\tan^{-1}\dfrac13+\tan^{-1}\dfrac18=\dfrac\pi4\).

Hint:

Pair the terms and repeatedly apply the tan inverse addition formula; the final ratio collapses to 1.

Answer 1

Step 1: Combine the first pair using \(\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}\):
\[ \tan^{-1}\dfrac15+\tan^{-1}\dfrac17=\tan^{-1}\dfrac{\frac15+\frac17}{1-\frac15\cdot\frac17}=\tan^{-1}\dfrac{12/35}{34/35}=\tan^{-1}\dfrac{12}{34}=\tan^{-1}\dfrac6{17} \]

Step 2: Combine the second pair the same way:
\[ \tan^{-1}\dfrac13+\tan^{-1}\dfrac18=\tan^{-1}\dfrac{\frac13+\frac18}{1-\frac13\cdot\frac18}=\tan^{-1}\dfrac{11/24}{23/24}=\tan^{-1}\dfrac{11}{23} \]

Step 3: Combine the two results:
\[ \tan^{-1}\dfrac6{17}+\tan^{-1}\dfrac{11}{23}=\tan^{-1}\dfrac{\frac6{17}+\frac{11}{23}}{1-\frac6{17}\cdot\frac{11}{23}} \]
Numerator: \(\dfrac{6\times23+11\times17}{17\times23}=\dfrac{138+187}{391}=\dfrac{325}{391}\).
Denominator: \(1-\dfrac{66}{391}=\dfrac{325}{391}\).
So the fraction is \(\dfrac{325/391}{325/391}=1\).

Step 4: Conclude:
\[ \tan^{-1}(1)=\dfrac\pi4 \]

Final Answer:
\[ \boxed{\tan^{-1}\tfrac15+\tan^{-1}\tfrac17+\tan^{-1}\tfrac13+\tan^{-1}\tfrac18=\dfrac\pi4} \]