Difference between revisions of "1985 AHSME Problems/Problem 16"

Revision as of 11:48, 25 January 2023

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

Solution 3

Let's write out the expression in terms of sine and cosine, so that we may see that it is equal to $\left(1+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(1+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = \left(\frac{\cos 20^\circ}{\cos 20^\circ}+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(\frac{\cos 25^\circ}{\cos 25^\circ}+\frac{\sin 25^\circ}{\cos 25^\circ}\right) =$ $\frac{\cos 20^\circ \cos 25^\circ + \cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.$ Clearly, that is equal to $1 + \frac{\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.$ Now, we note that $\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ$ is equal to $\sin 45^\circ$ . Now, we would like to get $\sin 20^\circ \sin 25^\circ$ in the denominator. What springs to mind is the fact that $\cos 20^\circ \cos 25^\circ - \sin 20^\circ \sin 25^\circ = \cos 45^\circ.$ Therefore, we can express the desired value as $1 + \frac{\sin 45^\circ + \sin 20^\circ \sin 25^\circ}{\cos 45^\circ + \sin 20^\circ \sin 25^\circ}.$ Because $\cos 45^\circ = \sin 45^\circ$ , we see that the fractional part is $1$ , and so the sum is $1 + 1 = 2$ , which brings us to the answer $\boxed{B}$ .

Solution 4

Similar to above: $\left(\frac{\cos A+\sin A}{\cos A}\right)\left(\frac{\cos B+\sin B}{\cos B}\right)$

Notice that $(\cos A + \sin A)^2 = \cos ^2 A + \sin ^2 A + 2\sin A \cos A = 1 + \sin (2A)$

Thus, $\left(\frac{\sqrt{1+\sin (2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin (2B)}}{\cos B}\right)$ $= 2\left(\frac{\sqrt{\frac{1+\cos (90 -2A)}{2}}}{\cos A}\right)\left(\frac{\sqrt{\frac{1+\cos (90-2B)}{2}}}{\cos B}\right)$

By half angle identity, $2\left(\frac{\cos (45 - A)}{\cos A}\right)\left(\frac{\cos (45 - B)}{\cos B}\right) = 2\frac{\cos 25 \cos 20}{\cos 20 \cos 25} = \hbox{\text{(B) } 2}$

@@ Line 76: / Line 76: @@
 Thus, <math> \left(\frac{\sqrt{1+\sin (2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin (2B)}}{\cos B}\right)</math> <math> = 2\left(\frac{\sqrt{\frac{1+\cos (90 -2A)}{2}}}{\cos A}\right)\left(\frac{\sqrt{\frac{1+\cos (90-2B)}{2}}}{\cos B}\right)</math>
-By half angle identity, <math> 2\left(\frac{\cos (45 - A)}{\cos A}\right)\left(\frac{\cos (45 - B)}{\cos B}\right) = 2\frac{\cos 25 \cos 20}{\cos 20 \cos 25} = 2</math>
+By half angle identity, <math> 2\left(\frac{\cos (45 - A)}{\cos A}\right)\left(\frac{\cos (45 - B)}{\cos B}\right) = 2\frac{\cos 25 \cos 20}{\cos 20 \cos 25} = \hbox{\text{(B) } 2}</math>
 ==See Also==
 {{AHSME box|year=1985|num-b=15|num-a=17}}
 {{MAA Notice}}

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