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

1. ## 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

2. 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.

3. 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 ${\color{red}-} \frac{13}{12}$.

4. 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
$\sec\left(\arctan(x)\right)=\sqrt{x^2+1}$

5. 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.

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If y = arctan(5/(-12)),
Then, opp = 5, adj = -12, hyp = 13

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