Question 840926
Find the value of sec(θ) if tan(θ)= {{{-sqrt(114)/19}}} and sin(θ)>0.
<pre>
There are two ways to do this problem.  We'll do it both ways:

First way.  &#952; is in quadrant II because that 
is the only quadrant in which the tangent 
is a negative number and the sine is a 
positive number.

So we draw the angle &#952; in quadrant II.
The tangent = {{{(opposite)/(adjacent)}}} = {{{y/x}}}. So
we take the numerator of the tangent {{{""+sqrt(114)}}} as y, 
(we take it positive because it goes up from the x-axis).
And we take the denominator of the tangent -19 as x, 
(we take it negative because it goes left from the origin).

We use the Pythagorean theorem: {{{hypotenuse^2=adjacent^2+opposite^2}}}
or rē = xē + yē
   rē = 19ē + {{{(sqrt(114))^2}}}
   rē = 361 + 114
   rē = 475
    r = {{{sqrt(475)}}} = {{{sqrt(25*19)}}} = {{{5sqrt(19)}}}

(r is always taken positive).

{{{drawing(500,13000/33,-25,8,-13,13,graph(500,13000/33,-25,8,-13,13),
line(-19,0,-19,sqrt(114)),line(0,0,-19,sqrt(114)),
locate(-13 , 1.6, "x=-19" ),locate(-23,5.3,y=sqrt(114)),
locate(-10,7,r=5sqrt(19)), red(arc(0,0,7,-7,0,150),locate(1,4.5,theta)) 



  )}}}

And since we know that the secant = {{{hypotenuse/adjacent}}} = {{{r/x}}},

sec(&#952;) = {{{(5sqrt(19))/(-19)}}} = {{{-5sqrt(19)/19}}}

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

Second way:

Use the identity: 

1 + tanē(&#952;) = secē(&#952;)
1 + {{{(-sqrt(114)/19)^2}}} = secē(&#952;)
1 + {{{114/361}}} = secē(&#952;)
{{{361/361}}} + {{{114/361}}} = secē(&#952;)
{{{475/361}}} = secē(&#952;)
{{{"" +- sqrt(475/361)}}} = sec(&#952;)
{{{"" +- sqrt(475)/sqrt(361)}}} = sec(&#952;) 
{{{"" +- sqrt(25*19)/19}}} = sec(&#952;)
{{{"" +- 5sqrt(19)/19}}} = sec(&#952;)

We have to decide which quadrant &#952; is in just
as we did using the first method.  So the secant
in QII is negative, so the answer is

{{{- 5sqrt(19)/19}}} = sec(&#952;)

You can do the problem either way.

Edwin</pre>