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

Nov 19, 2019

#1
+3

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!

!    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
+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

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

Thanks asinus and CPhill,