Hey guys i want to know if i am correct here

let y=arctan(-5/12)

sec y= 13/12 or 65 degrees

is this correct

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- Jun 30th 2008, 08:02 PMNoAsherelolFInd exact value of the expression Sec(arctan -5/12)
Hey guys i want to know if i am correct here

let y=arctan(-5/12)

sec y= 13/12 or 65 degrees

is this correct - Jun 30th 2008, 08:56 PMo_O
Yep, except you might not want the 65 degrees as your answer as that is obviously

**not**an exact answer and is not even equivalent to your expression.

$\displaystyle \sec\left[ \arctan \left( -\frac{5}{12}\right) \right]$ should be unitless. - Jun 30th 2008, 09:11 PMmr fantastic
- Jun 30th 2008, 09:22 PMMathstud28
- Jul 1st 2008, 02:45 AMticbol
If y = arctan(-5 /12),

then, tan(y) = -5/ 12

So, y is an angle whose tan value is -5 /12.

tan = (opposite side)/(adjacent side) = opp/adj

So, for angle y above,

opp = -5

adj = 12

The hypotenuse---hyp---then is sqrt[(-5)^2 +(12)^2] = 13

sec = hyp/adj

So, sec(y) = 13/12 -----you got that right.

-----------------------

If y = arctan(5/(-12)),

Then, opp = 5, adj = -12, hyp = 13

sec(y) = 13/(-12) or -(13/12)