Suppose that \(\sec x+\tan x=\frac{22}7. \) Find \(\csc x+\cot x. \)
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\(\color{BrickRed}\sec x+\tan x=\frac{22}7\\ \pm\sqrt{1+tan^2x}+\tan x=\frac{22}7\\ \pm\sqrt{1+tan^2x}=\frac{22}7-tan\ x\\ 1+tan^2x=\frac{484}{49}-\frac{44}{7}tan\ x+tan^2x\\ \frac{44}{7}tan\ x=\frac{484}{49}-1\)
\(\frac{44}{7}tan\ x=\frac{435}{49}\\ tan\ x=\frac{3\cdot 5\cdot 29\cdot 7}{7\cdot 7\cdot 4\cdot 11}=\frac{3\cdot 5\cdot 29}{4\cdot 7\cdot 11}\)
\(tan\ x=\frac{435}{308}\)
\(x=0.954690764748\)
\(csc\ x=\frac{533}{435}\)
\(cot\ x =\frac{308}{435}\)
\(csc\ x+ cot\ x=\frac{841}{435}\)
\(\large csc\ x+ cot\ x=\frac{29}{15}\)
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+128460 sec x + tan x = 22/7 square both sides
sec^2 x + 2sec x tan x + tan^2 x = 484/49
sec^2 x + 2sec x tan x + (sec^2 x - 1) = 484 / 49
2sec^2x + 2sec x tanx = 484/49 + 1
2secx ( sec x + tan x) = 533/ 49
2 sec x ( 22/7) = 533/49
(44/7) sec x = 533/49 multiply both sides by 7/44
sec x = 533/308
So cos x = 308/533
And we can solve for the sin as follows
sec x + tan x = 22/7
533/308 + sin x / ( 308/533) = 22/7
533/308 + (533/308)sinx = 22/7
(533/308) sin x = 22/7 - 533/308
(533/308) sin x = 435 / 308
sin x = 435 / 533
So
csc x + cot x =
1 / sin x + cos x /sin x =
1/ [ 435/533 ] + (308 /533) / (435/533) =
533/ 435 + 308 / 435 =
831 / 435 =
(29 * 29) / (29 * 15) =
29/15
