1985 AHSME Problems/Problem 16

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 $\begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= 2\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2A\right)}{2}}}{\cos A}\right)\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2B\right)}{2}}}{\cos B}\right) \\ &= 2\left(\frac{\cos\left(45^{\circ}-A\right)}{\cos A}\right)\left(\frac{\cos\left(45^{\circ}-B\right)}{\cos B}\right) \\ &= 2\left(\frac{\cos 25^{\circ} \cos 20^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}\right) = \boxed{\text{(B)} \ 2}.\end{align*}$

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