1991 AIME Problems/Problem 9
Suppose that and that where is in lowest terms. Find
Use the two trigonometric Pythagorean identities and .
If we square the given , we find that
This yields .
Let . Then squaring,
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 = , and then computing csc x + cot x using 1/sin x and then the reciprocal of tan x.
Recall that , from which we find that . Adding the equations
together and dividing by 2 gives , and subtracting the equations and dividing by 2 gives . Hence, and . Thus, and . Finally,
Solution 3 (least computation)
By the given, and .
Multiplying the two, we have
Subtracting both of the two given equations from this, and simpliyfing with the identity , we get
Solving yields , and
Make the substitution (a substitution commonly used in calculus). By the half-angle identity for tangent, , so . Also, we have Now note the following:
Plugging these into our equality gives:
This simplifies to , and solving for gives , and . Finally, .
We are given that , or equivalently, . Note that what we want is just .
Assign a right triangle with angle , hypotenuse , adjacent side , and opposite side . Then, through the given information above, we have that..
Hence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms.
Furthermore, by the Pythagorean Theorem, we have that
Solving for in the first equation and plugging in into the second equation...
Now, we want
Plugging in, we find the answer is
Hence, the answer is
We know that sec(x) = and that tan(x) = where , , represent the hypotenuse, adjacent, and opposite (respectively) to angle x in a right triangle. Thus we have that sec(x) + tan(x) = . We also have that csc(x) + cot(x) = . Set sec(x) + tan(x) = and csc(x)+cot(x) = . Then, notice that ( This is because of the Pythagorean Theorem, recall ). But then notice that . From the information provided in the question, we can substitute for . Thus, . Since, essentially we are asked to find the sum of the numerator and denominator of , we have .
|1991 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|