Question 1118984
The drawing below shows the situation for Q1  (ignore the labeling on the x and y axes… the triangles are not to scale).<br>

{{{ drawing(400,400, -5, 5, -2, 10, grid(0),
line(0,0, 4,0),
line(4,0,4,8.5),
line(0,0,4,8.5),
locate(4.3,4.2, "8"),
locate(2,-0.4, "x=15"),
locate(1.8, 5.6, "17"),
locate(0.3,0.5,theta)
)
}}}

{{{ x = sqrt(17^2 - 8^2) = sqrt(225) = 15 }}} 

{{{ tan(theta) = 8/15 }}}
{{{ sec(theta) = 1/cos(theta) = 1/(15/17) = 17/15 }}}

{{{ tan(theta) + sec(theta) = 8/15 + 17/15 = 25/15 = highlight( 5/3 ) }}}

The other angle is in Q2 (where {{{ sin(pi-x) = sin(x) }}} ) so the sine value is still 8/17  (i.e. positive) but the tan & sec values are both negative, resulting in   {{{ highlight(-5/3) }}}   To visualize, flip the triangle across the y-axis where  x<0,  and notice {{{theta}}} is still measured counter-clockwise from the positive x-axis but now ends in Q2:<br>

{{{ drawing(400,400, -5, 5, -2, 10, grid(0),
line(0,0, -4,0),
line(-4,0,-4,8.5),
line(0,0,-4,8.5),
locate(-4.3,4.2, "8"),
locate(-2,-0.4, "x= -15"),
locate(-1.8, 5.4, "17"),
locate(0.3,0.5,theta)
)
}}}