Question

Use the method of Example 6 to prove the identity.

38. \(\arctan x + \arctan \left( {\frac{1}{x}} \right) = \frac{\pi }{2}\), \(x > 0\)

Short answer

The identity has been proved.

Step-by-step solution

Write Theorem 5

If \(f'\left( x \right) = 0\) for all \(x\) in an interval \(\left( {a,b} \right)\), then \(f\) is constanton \(\left( {a,b} \right)\).

Proof the identity

Let a function \(f\left( x \right) = \arctan x + \arctan \left( {\frac{1}{x}} \right)\).

Differentiate both the sides with respect to \(x\) as follows:

\(\begin{aligned}{c}f'\left( x \right) &= \frac{d}{{dx}}\left( {\arctan x} \right) + \frac{d}{{dx}}\left( {\arctan \left( {\frac{1}{x}} \right)} \right)\\ &= \frac{1}{{1 + {x^2}}} + \frac{1}{{1 + \frac{1}{{{x^2}}}}} \cdot \frac{d}{{dx}}\left( {\frac{1}{x}} \right)\\ &= \frac{1}{{1 + {x^2}}} + \frac{{{x^2}}}{{{x^2} + 1}} \cdot \left( { - \frac{1}{{{x^2}}}} \right)\\ &= \frac{1}{{1 + {x^2}}} - \frac{1}{{{x^2} + 1}}\\ &= 0\end{aligned}\)

Here, \(f'\left( x \right) = 0\). Therefore, by theorem 5, \(f\left( x \right) = c\) where \(c\) is a constant.

Find the value of c

Here, \(f\left( x \right) = \arctan x + \arctan \left( {\frac{1}{x}} \right)\). Let \(x = 1\), then \(\arctan 1 + \arctan \frac{1}{1} = c\). That implies \(\frac{\pi }{4} + \frac{\pi }{4} = c \Rightarrow c = \frac{\pi }{2}\).

Hence, it is proved that \(\arctan x + \arctan \left( {\frac{1}{x}} \right) = \frac{\pi }{4}\).