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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

Solution 1

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 $n$.

Solution 2

We can see that $25^o+20^o=45^o$. We also know that $\tan 45=1$. First, let us expand $(1+\tan A)(1+\tan B)$.

We get $1+\tan A+\tan B+\tan A\tan B$.

Now, let us look at $\tan45=\tan(20+25)$.

By the $\tan$ sum formula, we know that $\tan45=\dfrac{\text{tan A}+\text{tan B}}{1- \text{tan A} \text{tan B}}$

Then, since $\tan 45=1$, we can see that $\tan A+\tan B=1-\tan A\tan B$

Then $1=\tan A+\tan B+\tan A\tan B$

Thus, the sum become $1+1=2$ and the answer is $\fbox{\text{(B)}}$

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