1979 AHSME Problems/Problem 20

Problem #20

If $a=\tfrac{1}{2}$ and $(a+1)(b+1)=2$ then the radian measure of $\arctan a + \arctan b$ equals

$\textbf{(A) }\frac{\pi}{2}\qquad \textbf{(B) }\frac{\pi}{3}\qquad \textbf{(C) }\frac{\pi}{4}\qquad \textbf{(D) }\frac{\pi}{5}\qquad \textbf{(E) }\frac{\pi}{6}$

SOLUTION

Solution by e_power_pi_times_i

Since $a=\frac{1}{2}$ , $b=\frac{1}{3}$ . Now we evaluate $\arctan a$ and $\arctan b$ . Denote $x$ and $\theta$ such that $\arctan x = \theta$ . Then $\tan(\arctan(x)) = \tan(\theta)$ , and simplifying gives $x = \tan(\theta)$ . So $a = \tan(\theta_a) = \frac{1}{2}$ and $b = \tan(\theta_b) = \frac{1}{3}$ . The question asks for $\theta_a + \theta_b$ , so we try to find $\tan(\theta_a + \theta_b)$ in terms of $\tan(\theta_a)$ and $\tan(\theta_b)$ . Using the angle addition formula for $\tan(\alpha+\beta)$ , we get that $\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}$ . Plugging $\tan(\theta_a) = \frac{1}{2}$ and $\tan(\theta_b) = \frac{1}{3}$ in, we have $\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}$ . Simplifying, $\tan(\theta_a + \theta_b) = 1$ , so $\theta_a + \theta_b$ in radians is $\boxed{\textbf{(C) } \frac{\pi}{4}}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1979 AHSME Problems/Problem 20

Problem #20

SOLUTION

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