1985 AHSME Problems/Problem 16

Problem

If $A=20^\circ$ and $B=25^\circ$ , then the value of $(1+\tan A)(1+\tan B)$ is

$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these}$

Solution

First, let's leave everything in variables and see if we can simplify $(1+\tan A)(1+\tan B)$ .

We can write everything in terms of sine and cosine to get $\left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right)=\frac{(\sin A+\cos A)(\sin B+\cos B)}{\cos A\cos B}$ .

We can multiply out the numerator to get $\frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B}$ .

It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:

$\cos(A-B)=\sin A\sin B+\cos A\cos B$

$\sin(A+B)=\sin A\cos B+\sin B\cos A$

Therefore, our fraction is equal to $\frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B}$ .

We can also use the product-to-sum formula

$\cos A\cos B=\frac{1}{2}(\cos(A-B)+\cos(A+B))$ to simplify the denominator:

$\frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\cos(A+B))}$ .

But now we seem stuck. However, we can note that since $A+B=45^\circ$ , we have $\sin(A+B)=\cos(A+B)$ , so we get

$\frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\sin(A+B))}$

$\frac{1}{\frac{1}{2}}$

$2, \boxed{\text{B}}$

Note that we only used the fact that $\sin(A+B)=\cos(A+B)$ , so we have in fact not just shown that $(1+\tan A)(1+\tan B)=2$ for $A=20^\circ$ and $B=25^\circ$ , but for all $A, B$ such that $A+B=45^\circ+n180^\circ$ , for integer $n$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1985 AHSME Problems/Problem 16

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