University of South Carolina High School Math Contest/1993 Exam/Problem 18

Problem

The minimum value of the function

$\displaystyle f(x) = \frac{\sin (x)}{\sqrt{1 - \cos^2 (x)}} + \frac{\cos(x)}{\sqrt{1 - \sin^2 (x) }} + \frac{\tan(x)}{\sqrt{\sec^2 (x) - 1}} + \frac{\cot (x)}{\sqrt{\csc^2 (x) - 1}}$

as $x$ varies over all numbers in the largest possible domain of $f$ , is

$\mathrm{(A) \ }-4 \qquad \mathrm{(B) \ }-2 \qquad \mathrm{(C) \ }0 \qquad \mathrm{(D) \ }2 \qquad \mathrm{(E) \ }4$

Solution

Recall the trigonometric identities

$\sin^2 x + \cos^2 x = 1$
$\tan^2 x + 1 = \sec^2 x$
$1 + \cot^2 x = \csc^2 x$

Since $\sqrt{x^2} = |x|$ for real $x$ , we can now simplify the function to

$f(x) = \frac{\sin(x)}{|\sin (x)|}+\frac{\cos(x)}{|\cos(x)|} + \frac{\tan(x)}{|\tan(x)|} + \frac{\cot(x)}{|\cot(x)|}.$

Now we must consider the quadrant that $x$ is in. If $x$ is in quadrant I, then all of the trig functions are positive and $f(x)=1+1+1+1=4$ . If $x$ is in quadrant II, then sine is positive and the rest of cosine, tangent, and cotangent are negative giving $f(x)=1-1-1-1=-2$ . If $x$ is in quadrant III, then tangent and cotangent are positive while sine and cosine are negative, making $f(x)=1+1-1-1=0$ . Finally, if $x$ is in quadrant IV, then only cosine is positive with the other three being negative giving $f(x)=-1+1-1-1=-2$ . Thus our answer is -2.

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University of South Carolina High School Math Contest/1993 Exam/Problem 18

Problem

Solution