Difference between revisions of "1999 AHSME Problems/Problem 15"
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\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{\textbf{(E)}\ 0.5}</math>. | <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)== | ||
| + | Note that <math>\sec x - \tan x = (1-\sin x)/\cos x</math>, and we can say that <math>\sec x + \tan x = (1+\sin x)/\cos x</math>, and if we let that equal y and multiply the two, we get <math>(1-\sin^{2}x)/\cos^{2}x</math>, which equals <math>1</math>. This equates to <math>2y = 1</math>. Thus, \boxed{\textbf{(E)}\ 0.5}$. | ||
==See Also== | ==See Also== | ||
Revision as of 19:23, 1 May 2023
Problem
Let 
be a real number such that 
. Then 
Solution 1 (Fastest)
, so 
.
Solution 2 (Alternate)
Note that 
, and we can say that 
, and if we let that equal y and multiply the two, we get 
, which equals 
. This equates to 
. Thus, \boxed{\textbf{(E)}\ 0.5}$.
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. 
