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1985 AHSME Problems/Problem 16

Revision as of 21:35, 6 March 2012 by Freddylukai (talk | contribs) (→‎Alternate Solution)

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 $n$.


Alternate Solution

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 $\fbox{\text{(B)}}$

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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