Question

How do you find the exact value `sec(x-y)` if `cscx=5/3,tany=12/5`?

Answer 1

The answer is `=-65/16`

Explanation:

`sec(x-y)=1/cos(x-y)=1/(cosxcosy-sinxsiny)`

`cscx=1/sinx=5/3`

Therefore, `sinx=3/5`

`cos^2x+sin^2x=1`

`cos^2x=1-9/25=16/25`

`cosx=4/5`

`tany=12/5`

`tan^2y+1=1/cos^2y`

`1/cos^2y=1+144/25=169/25`

`cosy=5/13`

`sin^2y=1-cos^2y=1-25/169=144/169`

`siny=12/13`

`sec(x-y)=1/(cosxcosy-sinxsiny)`

`=1/(4/5*5/13-3/5*12/13)`

`=1/(4/13-36/65)`

`=-65/16`

Answer 2

`sec(x-y)=65/56` if `x,y in (0.pi/2)`

Explanation:

`sec(x-y) =1/cos(x-y)= 1/(cosxcosy+sinxsiny)`

`cscx =1/sinx=5/3 => sinx = 3/5`

`cosx =+-sqrt(1-(3/5)^2) = +-sqrt(1-9/25) =+- sqrt(16/25) =+-4/5`

`tany=12/5 => sin^2y/(1-sin^2y) = (12/5)^2`

`25sin^2y+144sin^2y-144 =0`

`169sin^2y=144`

`siny=+-12/13`

`cosy=sqrt(1-(12/13)^2)=+-5/13`

As the tangent is positive, `siny` and `cosy` have the same sign.

For the sake of simplicity let's assume `x` and `y` are in `(0,pi/2)`
so we take all the positive solutions:

`sinx=3/5`
`cosx=4/5`
`siny=12/13`
`cosy=5/13`

`sec(x-y) == 1/(4/5*5/13+3/5*12/13)=1/(20/65+36/65)=65/56`

Alternatively, `x` could be in the second quadrant, where:

`cosx=-4/5`

or `y` could be in the third quadrant where:
`siny=-12/13`
`cosy=-5/13`