If tanΘ=9/5, what is secΘ

Thanks Geno,

I do it the same as Geno but for me I must draw a triangle every time.

Then you can read the values of the triangle.  The only other thing you MUST consider is what quadrants it can be in and is the trig ratio going to be positive or negative or either.

If tanΘ=9/5, what is secΘ

tan is poitive so it must be in the 1st or third quadrant.

sec = 1/cos = hyp/adj   

cos and therefore sec will be positive in the 1st quad and negative in the 3rd quad. 

$$sec\theta = \frac{\pm\sqrt{106}}{5}$$

Our answers are different Geno, I think you have made a little mistake :)    $$sec\theta=\frac{1}{cos\theta}$$

When graphed in the usual x-y plane:  tan θ  =  y/x.

tan θ  =  y/x   and   tan θ  =  9/5 (if you assume that the angle is in the first quadrant).

The Pythagorean Theorem applies with  x² + y²  =  r²     --->   5² + 9²  =  r²

--->   25 + 81  =  r²     --->   106  =  r²

:  r  =  √106.

sec θ  =  1/ cos θ     and     cos θ  =  x/r     --->    cos θ  =  5/√106    

                                                              --->   sec θ  = √106/5.

However, the angle could be in the third quadrant which makes  y = -9  and  x = -5.

r is still √106, but  cos θ = -5/√106  and sec θ  = - √106/5.

This was corrected after notes from Melody and Alan -- thank you.