Difference between revisions of "1985 AHSME Problems/Problem 16"
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==Problem== | ==Problem== | ||
− | If <math> A=20 | + | If <math>\usepackage{gensymb} A = 20 \degree</math> and <math>\usepackage{gensymb} B = 25 \degree</math>, then the value of <math>\left(1+\tan A\right)\left(1+\tan B\right)</math> is |
− | <math> \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} </math> | + | <math> \mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these} </math> |
− | ==Solution== | + | ==Solution 1== |
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+ | Noting that <math>\usepackage{gensymb} 25 \degree + 20 \degree = 45 \degree</math>, we apply the angle sum formula <cmath>\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B},</cmath> giving <cmath> | ||
− | + | ==Solution 2== | |
+ | Expanding in terms of sines and cosines, we obtain <cmath>\begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= \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{\left(\sin A+\cos A\right)\left(\sin B+\cos B\right)}{\cos A\cos B} \ &= \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B}.\end{align*}</cmath> | ||
+ | Recalling the angle sum identities <cmath> | ||
+ | Now, using the product-to-sum formula <cmath>\cos A\cos B = \frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right),</cmath> we can simplify the denominator, yielding <cmath>\left(1+\tan A\right)\left(1+\tan B\right) = \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right)}.</cmath> | ||
− | + | Finally, since <math>\usepackage{gensymb} A+B = 45 \degree</math>, we have <math>\sin(A+B) = \cos(A+B)</math>, so | |
+ | <cmath>\begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\sin(A+B)\right)} \ &= \frac{1}{\left(\frac{1}{2}\right)} \ &= \boxed{\text{(B)} \ 2}.\end{align*}</cmath> | ||
+ | ''Remark'': Notice that we only used the fact that <math>\sin(A+B) = \cos(A+B)</math>, so we have in fact shown that <math>\left(1+\tan A\right)\left(1+\tan B\right) = 2</math> not just for <math>\usepackage{gensymb} A = 20 \degree</math> and <math>\usepackage{gensymb} B = 25 \degree</math>, but also for all <math>A,B</math> such that <math>\usepackage{gensymb} A+B = 45 \degree + 180 \degree n</math> for integers <math>n</math>. | ||
− | + | ==Solution 3== | |
+ | As in Solution 2, we rewrite the expression as <cmath>\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}},</cmath> and hence as <cmath>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}}.</cmath> Using the angle sum identities <cmath> | ||
− | + | ==Solution 4== | |
− | < | + | As in Solutions 2 and 3, the expression becomes <cmath>\left(\frac{\cos A+\sin A}{\cos A}\right)\left(\frac{\cos B+\sin B}{\cos B}\right).</cmath> |
+ | Now, using the identity <math>\cos^2 A + \sin^2 A = 1</math> and the double-angle identity <math>\sin(2A) = 2\sin A\cos A</math>, we observe that <cmath> | ||
− | + | Since <math>A</math> and <math>B</math> are acute, we have <math>\sin A,\cos A,\sin B,\cos B > 0</math>, so <math>\cos A + \sin A > 0</math> and <math>\cos B + \sin B > 0</math>. Hence, taking the positive square root of both sides in the above identity, the expression becomes <cmath>\left(\frac{\sqrt{1+\sin(2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin(2B)}}{\cos B}\right).</cmath> Recalling the further identity <math>\sin A = \cos\left(90^{\circ}-A\right)</math>, together with the half-angle identity <cmath>\cos\left(\frac{A}{2}\right) = \sqrt{\frac{1+\cos A}{2}} \qquad \text{for } 0^{\circ} \leq A \leq 180^{\circ},</cmath> we finally obtain <cmath>\begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= 2\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2A\right)}{2}}}{\cos A}\right)\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2B\right)}{2}}}{\cos B}\right) \ &= 2\left(\frac{\cos\left(45^{\circ}-A\right)}{\cos A}\right)\left(\frac{\cos\left(45^{\circ}-B\right)}{\cos B}\right) \ &= 2\left(\frac{\cos 25^{\circ} \cos 20^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}\right) = \boxed{\text{(B)} \ 2}.\end{align*}</cmath> | |
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==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=15|num-a=17}} | {{AHSME box|year=1985|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:25, 19 March 2024
Problem
If and , then the value of is
Solution 1
Noting that , we apply the angle sum formula giving so Hence
Solution 2
Expanding in terms of sines and cosines, we obtain
Recalling the angle sum identities this reduces to
Now, using the product-to-sum formula we can simplify the denominator, yielding
Finally, since , we have , so
Remark: Notice that we only used the fact that , so we have in fact shown that not just for and , but also for all such that for integers .
Solution 3
As in Solution 2, we rewrite the expression as and hence as Using the angle sum identities we obtain Therefore the expression becomes
Solution 4
As in Solutions 2 and 3, the expression becomes
Now, using the identity and the double-angle identity , we observe that
Since and are acute, we have , so and . Hence, taking the positive square root of both sides in the above identity, the expression becomes Recalling the further identity , together with the half-angle identity we finally obtain
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.