Question 1060297
<pre><font size = 4><b>
{{{sec(v - u)}}}{{{""=""}}}{{{1/cos(v-u)}}}{{{""=""}}}{{{1/(cos(v)cos(u)+sin(v)sin(u))}}}{{{""=""}}}{{{1[""]/((-3/5)cos(u)+sin(v)(5/13))}}}{{{""=""}}}

We have to pause here and use an identity:

{{{sin^2(theta)+cos^2(theta)}}}{{{""=""}}}{{{1}}}

we use it to find cos(u) from sin(u):

{{{sin^2(u)+cos^2(u)}}}{{{""=""}}}{{{1}}}

{{{(5/13)^2+cos^2(u)}}}{{{""=""}}}{{{1}}}

{{{25/169+cos^2(u)}}}{{{""=""}}}{{{1}}}

Multiply through by 169

{{{25+169cos^2(u)}}}{{{""=""}}}{{{169}}}

Subtract 25 from both sides:

{{{169cos^2(u)}}}{{{""=""}}}{{{144}}}

{{{cos^2(u)}}}{{{""=""}}}{{{144/169}}}

{{{cos(u)}}}{{{""=""}}}{{{"" +- sqrt(144/169)}}}

{{{cos(u)}}}{{{""=""}}}{{{"" +- 12/13}}}

We can determine which sign + or - to use because
we know that both are in Quadrant II, and the
cosine is negative in quadrant II.  So {{{cos(u)}}}{{{""=""}}}{{{-12/13}}}

we can now substitute that in 

{{{1[""]/((-3/5)cos(u)+sin(v)(5/13))}}}{{{""=""}}}

{{{1[""]/((-3/5)(-12/13)+sin(v)(5/13))}}}{{{""=""}}}

But we still need sin(v).  So we go back to the identity:

{{{sin^2(theta)+cos^2(theta)}}}{{{""=""}}}{{{1}}}

we use it to find sin(v) from cos(v):

{{{sin^2(v)+cos^2(v)}}}{{{""=""}}}{{{1}}}

{{{sin^2(v)+(-3/5)^2}}}{{{""=""}}}{{{1}}}  

{{{sin^2(v)+9/25}}}{{{""=""}}}{{{1}}}

Multiply through by 25

{{{25sin^2(v)+9}}}{{{""=""}}}{{{25}}}

Subtract 9 from both sides:

{{{25sin^2(v)}}}{{{""=""}}}{{{16}}}

{{{sin^2(v)}}}{{{""=""}}}{{{16/25}}}

{{{sin(v)}}}{{{""=""}}}{{{"" +- sqrt(16/25)}}}

{{{sin(v)}}}{{{""=""}}}{{{"" +- 4/5}}}

As before we can determine which sign + or - to use because
we know that both are in Quadrant II, and the sine is
positive in quadrant II.  So {{{sin(v)}}}{{{""=""}}}{{{"" + 4/5}}}

we can now substitute that in 


{{{1[""]/((-3/5)(-12/13)+(4/5)(5/13))}}}{{{""=""}}}{{{1[""]/((36/65)+(20/65))}}}{{{""=""}}}{{{(1[""])/(56/65)}}}{{{""=""}}}{{{1}}}{{{"÷"}}}{{{56/65}}}{{{""=""}}}{{{1}}}{{{""*""}}}{{{65/56}}}{{{""=""}}}{{{65/56}}}

Answer: {{{65/56}}}

Edwin</pre></b></font>