Difference between revisions of "1985 AHSME Problems/Problem 16"
(Created page with "==Problem== If <math> A=20^\circ </math> and <math> B=25^\circ </math>, then the value of <math> (1+\tan A)(1+\tan B) </math> is <math> \mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B...") |
Freddylukai (talk | contribs) (→Solution: added alternate solution) |
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Note that we only used the fact that <math> \sin(A+B)=\cos(A+B) </math>, so we have in fact not just shown that <math> (1+\tan A)(1+\tan B)=2 </math> for <math> A=20^\circ </math> and <math> B=25^\circ </math>, but for all <math> A, B </math> such that <math> A+B=45^\circ+n180^\circ </math>, for integer <math> n </math>. | Note that we only used the fact that <math> \sin(A+B)=\cos(A+B) </math>, so we have in fact not just shown that <math> (1+\tan A)(1+\tan B)=2 </math> for <math> A=20^\circ </math> and <math> B=25^\circ </math>, but for all <math> A, B </math> such that <math> A+B=45^\circ+n180^\circ </math>, for integer <math> n </math>. | ||
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+ | ==Alternate Solution == | ||
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+ | We can see that <math>25^o+20^o=45^o</math>. We also know that <math>\tan 45=1</math>. First, let us expand <math>(1+\tan A)(1+\tan B)</math>. | ||
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+ | We get <math>1+\tan A+\tan B+\tan A\tan B</math>. | ||
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+ | Now, let us look at <math>\tan45=\tan(20+25)</math>. | ||
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+ | By the <math>\tan</math> sum formula, we know that <math>\tan45=\dfrac{\text{tan A}+\text{tan B}}{1- \text{tan A} \text{tan B}}</math> | ||
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+ | Then, since <math>\tan 45=1</math>, we can see that <math>\tan A+\tan B=1-\tan A\tan B</math> | ||
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+ | Then <math>1=\tan A+\tan B+\tan A\tan B</math> | ||
+ | |||
+ | Thus, the sum become <math>1+1=2</math> and the answer is <math>\fbox{\text{(B)}}</math> | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=15|num-a=17}} | {{AHSME box|year=1985|num-b=15|num-a=17}} |
Revision as of 19:34, 9 February 2012
Contents
[hide]Problem
If and , then the value of is
Solution
First, let's leave everything in variables and see if we can simplify .
We can write everything in terms of sine and cosine to get .
We can multiply out the numerator to get .
It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:
Therefore, our fraction is equal to .
We can also use the product-to-sum formula
to simplify the denominator:
.
But now we seem stuck. However, we can note that since , we have , so we get
Note that we only used the fact that , so we have in fact not just shown that for and , but for all such that , for integer .
Alternate Solution
We can see that . We also know that . First, let us expand .
We get .
Now, let us look at .
By the sum formula, we know that
Then, since , we can see that
Then
Thus, the sum become and the answer is
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 |