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Suppose that \(\sec x+\tan x=\frac{22}7.\) Find \(\csc x+\cot x.\)

 Nov 19, 2019
 #1
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Suppose that \(\sec x+\tan x=\frac{22}7. \)   Find \(\csc x+\cot x. \)

 

Hello Guest!

 

\(\color{BrickRed}\sec x+\tan x=\frac{22}7\\ \pm\sqrt{1+tan^2x}+\tan x=\frac{22}7\\ \pm\sqrt{1+tan^2x}=\frac{22}7-tan\ x\\ 1+tan^2x=\frac{484}{49}-\frac{44}{7}tan\ x+tan^2x\\ \frac{44}{7}tan\ x=\frac{484}{49}-1\)

\(\frac{44}{7}tan\ x=\frac{435}{49}\\ tan\ x=\frac{3\cdot 5\cdot 29\cdot 7}{7\cdot 7\cdot 4\cdot 11}=\frac{3\cdot 5\cdot 29}{4\cdot 7\cdot 11}\)

\(tan\ x=\frac{435}{308}\)

 

\(x=0.954690764748\)

\(csc\ x=\frac{533}{435}\)

\(cot\ x =\frac{308}{435}\)

\(csc\ x+ cot\ x=\frac{841}{435}\)

\(\large csc\ x+ cot\ x=\frac{29}{15}\)

 

           Thank you heureka!

laugh  !    Not logged in, asinus.

 Nov 19, 2019
edited by asinus  Nov 19, 2019
edited by asinus  Nov 19, 2019
edited by asinus  Nov 21, 2019
 #3
avatar+106515 
+2

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

 

 

 

 

cool cool cool

 Nov 20, 2019
edited by CPhill  Nov 20, 2019
 #4
avatar+106885 
+1

Thanks asinus and CPhill,

I was wondering about this one.  laugh

 Nov 20, 2019

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