# FInd exact value of the expression Sec(arctan -5/12)

• June 30th 2008, 09:02 PM
NoAsherelol
FInd 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
• June 30th 2008, 09:56 PM
o_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.

$\sec\left[ \arctan \left( -\frac{5}{12}\right) \right]$ should be unitless.
• June 30th 2008, 10:11 PM
mr fantastic
Quote:

Originally Posted by NoAsherelol
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

The correct answer is NOT an angle .....

$\sec\left[ \arctan \left( -\frac{5}{12}\right) \right] = \frac{13}{12}$ OR ${\color{red}-} \frac{13}{12}$.
• June 30th 2008, 10:22 PM
Mathstud28
Quote:

Originally Posted by NoAsherelol
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

$\sec\left(\arctan(x)\right)=\sqrt{x^2+1}$
• July 1st 2008, 03:45 AM
ticbol
Quote:

Originally Posted by NoAsherelol
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

If y = arctan(-5 /12),
then, tan(y) = -5/ 12
So, y is an angle whose tan value is -5 /12.

So, for angle y above,
opp = -5
The hypotenuse---hyp---then is sqrt[(-5)^2 +(12)^2] = 13

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)