Difference between revisions of "1999 AHSME Problems/Problem 15"
Songmath20 (talk | contribs) (→Solution 2 (Alternate, Slightly Longer)) |
|||
| (15 intermediate revisions by 5 users not shown) | |||
| Line 9: | Line 9: | ||
\textbf{(E)}\ 0.5</math> | \textbf{(E)}\ 0.5</math> | ||
| − | ==Solution== | + | ==Solution 1 (Fastest)== |
| − | <math>(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1</math>, so <math>\sec x + \tan x = \boxed{\ | + | <math>(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1</math>, so <math>\sec x + \tan x = \boxed{\textbf{(E)}\ 0.5}</math>. |
| − | {{ | + | ==Solution 2 (Alternate, Slightly Longer)== |
| + | Note that <math>\sec x - \tan x = (1-\sin x)/\cos x</math> and <math>\sec x + \tan x = (1+\sin x)/\cos x</math>. Let <math>(1+\sin x)/\cos x = y</math>. Multiplying, we get <math>(1-\sin^{2}x)/\cos^{2}x = 1</math>.Then, <math>2y = 1</math>. <math>\sec x + \tan x = | ||
| + | \boxed{\textbf{(E)}\ 0.5}</math>. ~songmath20 | ||
| + | Edited 5.1.2023 | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1999|num-b=14|num-a=16}} | {{AHSME box|year=1999|num-b=14|num-a=16}} | ||
| + | |||
| + | [[Category:Introductory Trigonometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:36, 1 May 2023
Problem
Let 
be a real number such that 
. Then 
Solution 1 (Fastest)
, so 
.
Solution 2 (Alternate, Slightly Longer)
Note that 
and 
. Let 
. Multiplying, we get 
.Then, 
. 
. ~songmath20
Edited 5.1.2023
See Also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
