Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 4)
(→‎Solution 4)
Line 74: Line 74:
 
Notice that <math>(\cos A + \sin A)^2 = \cos ^2 A + \sin ^2 A + 2\sin A \cos A = 1 + \sin (2A)</math>
 
Notice that <math>(\cos A + \sin A)^2 = \cos ^2 A + \sin ^2 A + 2\sin A \cos A = 1 + \sin (2A)</math>
  
−
Thus, <math> \left(\frac{\sqrt{1+\sin (2A)}}{\cos A}\right)\left(\frac{1+\sin (2B)}{\cos B}\right) = 2\left(\frac{\frac{\sqrt{1+\cos (90 - 2A)}{2}}}{\cos A}\right)\left(\frac{\frac{1+\sin (90 - 2B)}{2}}{\cos B}\right)</math>
+
Thus, <math> \left(\frac{\sqrt{1+\sin (2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin (2B)}}{\cos B}\right) = 2\left(\frac{\frac{\sqrt{1+\cos (90 - 2A)}{2}}}{\cos A}\right)\left(\frac{\frac{1+\sin (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} = 2</math>

Revision as of 11:46, 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, $\left(\frac{\sqrt{1+\sin (2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin (2B)}}{\cos B}\right) = 2\left(\frac{\frac{\sqrt{1+\cos (90 - 2A)}{2}}}{\cos A}\right)\left(\frac{\frac{1+\sin (90 - 2B)}{2}}}{\cos B}\right)$ (Error compiling LaTeX. Unknown error_msg)

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_16&oldid=187904"