Question 1141239
<pre>
Obtuse means between 90° and 180°, exclusive of 90° and 180°.
So theta is in QII. so we draw an angle with terminal side in QII 
and assume the angle indicated by the red arc is theta:

{{{drawing(400,400,-5,5,-5,5, line(-6,0,6,0), line(0,-6,0,6),
line(0,0,-4,3), red(arc(0,0,2,-2,0,144)) )}}} 

Then we drop a perpendicular (in green) down to the x-axis from the end of the
terminal side of theta.  That makes a right triangle with the x-axis.

{{{drawing(400,400,-5,5,-5,5, line(-6,0,6,0), line(0,-6,0,6),
line(0,0,-4,3), red(arc(0,0,2,-2,0,144)),green(line(-4,3,-4,0)) )}}}

We know that the sine of theta is k, which is k/1.  We also know that
the sine is the opposite over the hypotenuse, sow we make the opposite
side (the green side) equal to the numerator of k/1, which is k, and we make the hypotenuse (the terminal side of theta) be 1.  Then we calculate the
adjacent side on the x-axis by the Pythagorean theorem:

{{{c^2=a^2+b^2}}}
{{{1^2=a^2+k^2}}}
{{{1=a^2+k^2}}}
{{{1-k^2=a^2}}}
{{{""+-sqrt(1-k^2)=a}}}
And since the adjacent side goe to the left of the origin we take
the negative square root,
{{{-sqrt(1-k^2)=a}}}


{{{drawing(400,400,-5,5,-5,5, line(-6,0,6,0), line(0,-6,0,6),
locate(-4.2,1.6,k),locate(-2.1,2,1), locate(-3.5,.8,-sqrt(1-k^2)),
line(0,0,-4,3), red(arc(0,0,2,-2,0,144)),green(line(-4,3,-4,0)) )}}}

Now let's draw the angle 90°+theta by adding 90° to theta, indicated
by the blue arrow, and we draw a perpendicular (in red) to the x-axis:

{{{drawing(400,400,-5,5,-5,5, line(-6,0,6,0), line(0,-6,0,6),
locate(-4.2,1.6,k),locate(-2.1,2,1), locate(-3.5,.8,-sqrt(1-k^2)),
line(0,0,-4,3), red(arc(0,0,2,-2,0,144)),green(line(-4,3,-4,0)),
line(0,0,-3,-4), blue(arc(0,0,2,-2,0,234)), red(line(-3,-4,-3,0))



 )}}}

The lower triangle is congruent to the upper one.  Sides that go left
or down are taken negative, so for 90°+theta we have:

{{{drawing(400,400,-5,5,-5,5, line(-6,0,6,0), line(0,-6,0,6),
locate(-2,0,-k), locate(-4.6,-1.4,-sqrt(1-k^2)),
locate(-1.4,-2,1),
line(0,0,-3,-4), blue(arc(0,0,2,-2,0,234)), red(line(-3,-4,-3,0))

)}}}

Finally, since the tangent is the opposite over the adjacent, we have

{{{tan(90^"o"+theta)=(-sqrt(1-k^2))/(-k^"")= sqrt(1-k^2)/k^""}}}

Edwin</pre>