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

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==Problem==
 
==Problem==
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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
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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
  
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<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>
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<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>
  
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==Solution==
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==Solution 1==
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===Solution 1===
 
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First, let's leave everything in variables and see if we can simplify <math> (1+\tan A)(1+\tan B) </math>.
 
  
 +
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>\begin{align*}1 &= \tan 45^{\circ} \\ &= \tan(A+B) \\ &= \frac{\tan A+\tan B}{1-\tan A\tan B},\end{align*}</cmath> so <cmath>\tan A + \tan B = 1-\tan A\tan B.</cmath> Hence <cmath>\begin{align*}(1+\tan A)(1+\tan B) &= 1+\tan A+\tan B+\tan A\tan B \\ &= 1+\left(1-\tan A\tan B\right)+\tan A\tan B \\ &= \boxed{\text{(B)} \ 2}.\end{align*}</cmath>
  
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We can write everything in terms of sine and cosine to get <math> \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} </math>.
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==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>\begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*}</cmath> this reduces to <cmath>\frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B}.</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>
  
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We can multiply out the numerator to get <math> \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B} </math>.
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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>.
  
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It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:
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==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>\begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*}</cmath> we obtain <cmath>\begin{align*}&\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} = \sin\left(20^{\circ}+25^{\circ}\right) = \sin 45^{\circ} \text{ and} \\ &\cos 20^{\circ} \cos 25^{\circ} - \sin 20^{\circ} \sin 25^{\circ} = \cos\left(20^{\circ}+25^{\circ}\right) = \cos 45^{\circ}.\end{align*}</cmath> Therefore the expression becomes <cmath>\begin{align*}1+\frac{\sin 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}} &= 1+1 \qquad \text{(since } \sin 45^{\circ} = \cos 45^{\circ}\text{)} \\ &= \boxed{\text{(B)} \ 2}.\end{align*}</cmath>
  
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<math> \cos(A-B)=\sin A\sin B+\cos A\cos B </math>
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==Solution 4==
  
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<math> \sin(A+B)=\sin A\cos B+\sin B\cos A </math>
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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>\begin{align*}\left(\cos A + \sin A\right)^2 &= \cos^2 A + \sin^2 A + 2\sin A \cos A \\ &= 1 + 2\sin A \cos A \\ &= 1 + \sin(2A).\end{align*}</cmath>
  
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Therefore, our fraction is equal to <math> \frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B} </math>.
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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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We can also use the product-to-sum formula
 
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<math> \cos A\cos B=\frac{1}{2}(\cos(A-B)+\cos(A+B)) </math> to simplify the denominator:
 
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<math> \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\cos(A+B))} </math>.
 
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But now we seem stuck. However, we can note that since <math> A+B=45^\circ </math>, we have <math> \sin(A+B)=\cos(A+B) </math>, so we get
 
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<math> \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\sin(A+B))} </math>
 
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<math> \frac{1}{\frac{1}{2}} </math>
 
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<math> 2, \boxed{\text{B}} </math>
 
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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>.
 
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===Solution 2===
 
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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>
 
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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}}
 +
{{MAA Notice}}

Latest revision as of 21:25, 19 March 2024

Problem

If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$, then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is

$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these}$

Solution 1

Noting that $\usepackage{gensymb} 25 \degree + 20 \degree = 45 \degree$, we apply the angle sum formula \[\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B},\] giving \begin{align*}1 &= \tan 45^{\circ} \\ &= \tan(A+B) \\ &= \frac{\tan A+\tan B}{1-\tan A\tan B},\end{align*} so \[\tan A + \tan B = 1-\tan A\tan B.\] Hence \begin{align*}(1+\tan A)(1+\tan B) &= 1+\tan A+\tan B+\tan A\tan B \\ &= 1+\left(1-\tan A\tan B\right)+\tan A\tan B \\ &= \boxed{\text{(B)} \ 2}.\end{align*}

Solution 2

Expanding in terms of sines and cosines, we obtain \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*}

Recalling the angle sum identities \begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*} this reduces to \[\frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B}.\]

Now, using the product-to-sum formula \[\cos A\cos B = \frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right),\] we can simplify the denominator, yielding \[\left(1+\tan A\right)\left(1+\tan B\right) = \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right)}.\]

Finally, since $\usepackage{gensymb} A+B = 45 \degree$, we have $\sin(A+B) = \cos(A+B)$, so \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*}

Remark: Notice that we only used the fact that $\sin(A+B) = \cos(A+B)$, so we have in fact shown that $\left(1+\tan A\right)\left(1+\tan B\right) = 2$ not just for $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$, but also for all $A,B$ such that $\usepackage{gensymb} A+B = 45 \degree + 180 \degree n$ for integers $n$.

Solution 3

As in Solution 2, we rewrite the expression as \[\frac{\cos 20^{\circ} \cos 25^{\circ} + \cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}},\] and hence as \[1 + \frac{\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}.\] Using the angle sum identities \begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*} we obtain \begin{align*}&\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} = \sin\left(20^{\circ}+25^{\circ}\right) = \sin 45^{\circ} \text{ and} \\ &\cos 20^{\circ} \cos 25^{\circ} - \sin 20^{\circ} \sin 25^{\circ} = \cos\left(20^{\circ}+25^{\circ}\right) = \cos 45^{\circ}.\end{align*} Therefore the expression becomes \begin{align*}1+\frac{\sin 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}} &= 1+1 \qquad \text{(since } \sin 45^{\circ} = \cos 45^{\circ}\text{)} \\ &= \boxed{\text{(B)} \ 2}.\end{align*}

Solution 4

As in Solutions 2 and 3, the expression becomes \[\left(\frac{\cos A+\sin A}{\cos A}\right)\left(\frac{\cos B+\sin B}{\cos B}\right).\]

Now, using the identity $\cos^2 A + \sin^2 A = 1$ and the double-angle identity $\sin(2A) = 2\sin A\cos A$, we observe that \begin{align*}\left(\cos A + \sin A\right)^2 &= \cos^2 A + \sin^2 A + 2\sin A \cos A \\ &= 1 + 2\sin A \cos A \\ &= 1 + \sin(2A).\end{align*}

Since $A$ and $B$ are acute, we have $\sin A,\cos A,\sin B,\cos B > 0$, so $\cos A + \sin A > 0$ and $\cos B + \sin B > 0$. Hence, taking the positive square root of both sides in the above identity, the expression becomes \[\left(\frac{\sqrt{1+\sin(2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin(2B)}}{\cos B}\right).\] Recalling the further identity $\sin A = \cos\left(90^{\circ}-A\right)$, together with the half-angle identity \[\cos\left(\frac{A}{2}\right) = \sqrt{\frac{1+\cos A}{2}} \qquad \text{for } 0^{\circ} \leq A \leq 180^{\circ},\] we finally obtain \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*}

See Also

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