1 Cscx Cotx Cosx Secx

Problem

Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in everyman terms. Detect $m+n^{}_{}.$

Solution 1

Apply the two trigonometric Pythagorean identities $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$ .

If we square the given $\sec x = \frac{22}{7} - \tan x$ , nosotros find that

$\begin{align*} \sec^2 x &= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\ 1 &= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}$

This yields $\tan x = \frac{435}{308}$ .

Let $y = \frac mn$ . Then squaring,

$\csc^2 x = (y - \cot x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.$

Substituting $\cot x = \frac{1}{\tan x} = \frac{308}{435}$ yields a quadratic equation: $0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)$ . It turns out that only the positive root will work, so the value of $y = \frac{29}{15}$ and $m + n = \boxed{044}$ .

Note: The problem is much easier computed if we consider what $\sec (x)$ is, then find the relationship between $\sin( x)$ and $cos (x)$ (using $\tan (x) = \frac{435}{308}$ , then computing $\csc x + \cot x$ using $1/\sin x$ so the reciprocal of $\tan x$ .

Solution 2

Recall that $\sec^2 x - \tan^2 x = 1$ , from which nosotros discover that $\sec x - \tan x = 7/22$ . Calculation the equations

$\begin{eqnarray*} \sec x + \tan x & = & 22/7 \\ \sec x - \tan x & = & 7/22\end{eqnarray*}$

together and dividing by 2 gives $\sec x = 533/308$ , and subtracting the equations and dividing by 2 gives $\tan x = 435/308$ . Hence, $\cos x = 308/533$ and $\sin x = \tan x \cos x = (435/308)(308/533) = 435/533$ . Thus, $\csc x = 533/435$ and $\cot x = 308/435$ . Finally,

$\csc x + \cot x = \frac {841}{435} = \frac {29}{15},$

so $m + n = 044$ .

Solution three (to the lowest degree computation)

By the given, $\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}$ and $\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k$ .

Multiplying the two, we accept

$\frac {1}{\sin x \cos x} + \frac {1}{\sin x} + \frac {1}{\cos x} + 1 = \frac {22}{7}k$

Subtracting both of the 2 given equations from this, and simpliyfing with the identity $\frac {\sin x}{\cos x} + \frac {\cos x}{\sin x} = \frac{\sin ^2 x + \cos ^2 x}{\sin x \cos x} = \frac {1}{\sin x \cos x}$ , nosotros get

$1 = \frac {22}{7}k - \frac {22}{7} - k.$

Solving yields $k = \frac {29}{15}$ , and $m+n = 044$

Solution 4

Brand the substitution $u = \tan \frac x2$ (a substitution normally used in calculus). By the half-bending identity for tangent, $\tan \frac x2 = \frac{\sin x}{1+\cos x}$ , and then $\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn$ . Too, nosotros accept $\sec x + \tan x = \frac{1 + \sin x}{\cos x}.$ Now note the following:

$\begin{align*}\sin x &= \frac{2u}{1+u^2}\\ \cos x &= \frac{1-u^2}{1+u^2}\end{align*}$

Plugging these into our equality gives:

$\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7$

This simplifies to $\frac{1+u}{1-u} = \frac{22}7$ , and solving for $u$ gives $u = \frac{15}{29}$ , and $\frac mn = \frac{29}{15}$ . Finally, $m+n = 044$ .

Solution five

We are given that $\frac{1+\sin x}{\cos x}=\frac{22}7\implies\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}$ $=\frac{\cos x}{1-\sin x}$ , or equivalently, $\cos x=\frac{7+7\sin x}{22}=\frac{22-22\sin x}7\implies\sin x=\frac{22^2-7^2}{22^2+7^2}$ $\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}$ . Note that what we want is just $\frac{1+\cos x}{\sin x}=\frac{1+\frac{2\cdot22\cdot7}{22^2+7^2}}{\frac{22^2-7^2}{22^2+7^2}}=\frac{22^2+7^2+2\cdot22\cdot7}{22^2-7^2}=\frac{(22+7)^2}{(22-7)(22+7)}=\frac{22+7}{22-7}$ $=\frac{29}{15}\implies m+n=29+15=\boxed{044}$ .

Solution vi

Assign a correct triangle with angle $x$ , hypotenuse $c$ , next side $a$ , and opposite side $b$ . Then, through the given information above, we take that..

$\frac{c}{a}+\frac{b}{a}=\frac{22}{7}\implies \frac{c+b}{a}=\frac{22}{7}$

$\frac{c}{b}+\frac{a}{b}=\frac{m}{n}\implies \frac{a+c}{b}=\frac{m}{n}$

Hence, because similar right triangles can be scaled upwards by a gene, we tin assume that this detail right triangle is indeed in simplest terms.

Hence, $a=7$ , $b+c=22$

Furthermore, by the Pythagorean Theorem, we have that

$a^2+b^2=c^2\implies 49+b^2=c^2$

Solving for $c$ in the start equation and plugging in into the 2nd equation...

$49+b^2=(22-b)^2\implies 49+b^2=484-44b+b^2\implies 44b=435\implies b=\frac{435}{44}$

Hence, $c=22-\frac{435}{44}=\frac{533}{44}$

Now, we want $\frac{a+c}{b}$

Plugging in, nosotros find the answer is $\frac{\frac{7\cdot{44}}{44}+\frac{533}{44}}{\frac{435}{44}}=\frac{841}{435}=\frac{29}{15}$

Hence, the answer is $29+15=\boxed{044}$

Solution 7

We know that $\sec(x) = \frac{h}{a}$ and that $\tan(x) = \frac{o}{a}$ where $h$ , $a$ , $o$ represent the hypotenuse, next, and contrary (respectively) to angle $x$ in a right triangle. Thus we take that $\sec(x) + \tan(x) = \frac{h+o}{a}$ . We also have that $\csc(x) + \cot(x) = \frac{h}{o} + \frac{a}{o} = \frac{h+a}{o}$ . Set $\sec(x) + \tan(x) = \alpha$ and csc(x)+cot(x) = $\beta$ . And so, notice that $\alpha + \beta = \frac{h+o}{a} + \frac{h+a}{o} = \frac{oh+ah+o^2 + a^2}{oa} = \frac{h(o+a+h)}{oa}$ ( This is because of the Pythagorean Theorem, recollect $o^2 +a^2 = h^2$ ). Only then detect that $\alpha \cdot \beta = \frac{(o+h)(a+h)}{oa} = \frac{oa +oh +ha +h^2}{oa} = 1+ \frac{h(o+a+h)}{oa} = 1+ \alpha + \beta$ . From the information provided in the question, we can substitute $\alpha$ for $\frac{22}{7}$ . Thus, $\frac{22 \beta}{7}= \beta + \frac{29}{7} \Longrightarrow 22 \beta = 7 \beta + 29 \Longrightarrow 15 \beta = 29 \Longrightarrow \beta = \frac{29}{15}$ . Since, essentially nosotros are asked to find the sum of the numerator and denominator of $\beta$ , we have $29 + 15 = \boxed{044}$ .

~qwertysri987

Solution viii

Firstly, we write $\sec x+\tan x=a/b$ where $a=22$ and $b=7$ . This volition allow united states of america to spot factorable expressions later. Now, since $\sec^2x-\tan^2x=1$ , this gives us $\sec x-\tan x=\frac{b}{a}$ Adding this to our original expressions gives the states $2\sec x=\frac{a^2+b^2}{ab}$ or $\cos x=\frac{2ab}{a^2+b^2}$ At present since $\sin^2x+\cos^2x=1$ , $\sin x=\sqrt{1-\cos^2x}$ And so we can write $\sin x=\sqrt{1-\frac{4a^2b^2}{(a^2+b^2)^2}}$ Upon simplification, we become $\sin x=\frac{a^2-b^2}{a^2+b^2}$ We are asked to find $1/\sin x+\cos x/\sin x$ then we tin write that every bit $\csc x+\cot x=\frac{1}{\sin x}+\frac{\cos x}{\sin x}$ $\csc x+\cot x=\frac{a^2+b^2}{a^2-b^2}+\frac{2ab}{a^2+b^2}\frac{a^2+b^2}{a^2-b^2}$ $\csc x+\cot x=\frac{a^2+b^2+2ab}{a^2-b^2}$ $\csc x+\cot x=\frac{(a+b)^2}{(a-b)(a+b)}$ $\csc x+\cot x=\frac{a+b}{a-b}$ Now using the fact that $a=22$ and $b=7$ yields, $\csc x+\cot x=\frac{29}{15}=\frac{p}{q}$ and then $p+q=15+29=\boxed{44}$

~Chessmaster20000

See too

1991 AIME (Problems • Answer Key • Resources)
Preceded past Trouble 8	Followed past Problem 10
i • 2 • three • 4 • v • vi • 7 • 8 • ix • 10 • 11 • 12 • 13 • xiv • 15
All AIME Issues and Solutions

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