Use the fact that the co/sine functions are [tex]2\pi[/tex]-periodic and that the tangent function is [tex]\pi[/tex]-periodic. Also, recall that [tex]\cos x[/tex] is even (so that [tex]\cos(-x)=\cos x[/tex]) and [tex]\sin x[/tex] is odd (so that [tex]\sin(-x)=-\sin x[/tex].
15.
[tex]\cos\left(-\dfrac{13\pi}3\right)=\cos\dfrac{13\pi}3=\cos\left(\dfrac\pi3+4\pi\right)=\cos\dfrac\pi3=\boxed{\dfrac12}[/tex]
16.
[tex]\sin\dfrac{23\pi}4=\sin\left(\dfrac{3\pi}4+5\pi\right)=\sin\left(\dfrac{3\pi}4+\pi\right)=\sin\dfrac{7\pi}4=-\dfrac1{\sqrt2}[/tex]
[tex]\implies\csc\dfrac{23\pi}4=\boxed{-\sqrt2}[/tex]
17.
[tex]\cos\left(-\dfrac{7\pi}2\right)=\cos\dfrac{7\pi}2=\cos\left(\dfrac\pi2+3\pi\right)=\cos\left(\dfrac\pi2+\pi\right)=\cos\dfrac{3\pi}2=0[/tex]
[tex]\implies\sec\left(-\dfrac{7\pi}2\right)=\boxed{\text{undefined}}[/tex]
18.
[tex]\tan\left(-\dfrac{29\pi}6\right)=\dfrac{\sin\left(-\frac{29\pi}6\right)}{\cos\left(-\frac{29\pi}6\right)}=-\dfrac{\sin\frac{29\pi}6}{\cos\frac{29\pi}6}[/tex]
[tex]\sin\dfrac{29\pi}6=\sin\left(\dfrac{5\pi}6+4\pi\right)=\sin\dfrac{5\pi}6=-\dfrac12[/tex]
[tex]\cos\dfrac{29\pi}6=\cos\dfrac{5\pi}6=\dfrac{\sqrt3}2[/tex]
[tex]\implies\tan\left(-\dfrac{29\pi}6\right)=-\dfrac{-\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}[/tex]
[tex]\implies\cot\left(-\dfrac{29\pi}6\right)=\boxed{\sqrt3}[/tex]
