$\cos \theta =-0/8\xrightarrow{\theta }\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-{{(0/8)}^{2}}}=\sqrt{1-0/64}\Rightarrow \sin \theta =\sqrt{0/36}=0/6$ $\tan \theta -\sin (\theta -\frac{3\pi }{2})=\frac{\sin \theta }{\cos \theta }-(-\sin (\frac{3\pi }{2}-\theta ))=\frac{\sin \theta }{\cos \theta }+\sin (\frac{3\pi }{2}-\theta )$  و بنابراین داریم: $=\frac{\sin \theta }{\cos \theta }-\cos \theta =\frac{0/6}{-0/8}-(-0/8)=-\frac{6}{8}+0/8=-\frac{3}{4}+0/8=-0/75+0/8=0/05$