I guess you have to find
[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}[/tex]
given that [tex]\tan\theta=\frac8{15}[/tex].
We can immediately solve for [tex]\sec\theta[/tex]:
[tex]\sec^2\theta=1+\tan^2\theta\implies\sec\theta=\pm\dfrac{17}{15}[/tex]
(without knowing anything else about [tex]\theta[/tex], we cannot determine the sign)
Then we get [tex]\cos\theta[/tex] for free:
[tex]\cos\theta=\dfrac1{\sec\theta}=\pm\dfrac{15}{17}[/tex]
and we can now solve for [tex]\sin\theta[/tex]:
[tex]\sin^2\theta+\cos^2\theta=1\implies \sin\theta=\pm\dfrac8{17}[/tex]
Notice that we have 2*2 = 4 possible choices of sign for either sin or cos.
• If both are positive, then
[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}=\dfrac{391}{90}[/tex]
• If both are negative, then
[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}=\dfrac{391}{480}[/tex]
• If sin is positive and cos is negative, then
[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}=\dfrac{119}{480}[/tex]
• If cos is positive and sin is negative, then
[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}=\dfrac{119}{30}[/tex]
