Difference between revisions of "1999 AHSME Problems/Problem 15"
Songmath20 (talk | contribs) (→Solution 2 (Alternate)) |
Songmath20 (talk | contribs) (→Solution 2 (Alternate, Slightly Longer)) |
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==Solution 2 (Alternate, Slightly Longer)== | ==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> | + | 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>\sec x + \tan 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 | \boxed{\textbf{(E)}\ 0.5}</math>. ~songmath20 | ||
Edited 5.1.2023 | Edited 5.1.2023 |
Revision as of 18:34, 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 | |
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All AHSME Problems and Solutions |
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