Right away,
[tex]\sec\theta=-\dfrac{15}9\implies\cos\theta=-\dfrac9{15}=-\dfrac35[/tex]
Recall that [tex]\sin^2\theta+\cos^2\theta=1[/tex], so [tex]\sin\theta=\pm\sqrt{1-\cos^2\theta}[/tex]. We're told that [tex]\dfrac\pi2\le\theta\le\pi[/tex], for which we expect to have [tex]\sin\theta>0[/tex] and so we take the positive square root. Also recall that 3-4-5 is a Pythagorean triple. Then
[tex]\sin\theta=\dfrac45[/tex]
We have everything we need to find the remaining trig ratios. With the help of double angle identities, we get
[tex]\sin2\theta=2\sin\theta\cos\theta=2\left(\dfrac45\right)\left(-\dfrac35\right)=-\dfrac{24}{25}[/tex]
[tex]\cos2\theta=\cos^2\theta-\sin^2\theta=\left(-\dfrac35\right)^2-\left(\dfrac45\right)^2=-\dfrac7{25}[/tex]
[tex]\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{24}7[/tex]
