1.

`(sin(pi+x)cos(pi/2+x)tan((3pi)/2-x)cot(2pi-x))/(sin(2pi-x)cos(2pi+x)c o s e c(-x)sin((3pi)/2-x))=1`

Answer» `=(sin(pi+x)cos(pi/2+x)tan(3/2pi-x)cot(2pi-x))/(sin(2pi-x)cos(2pi+x)cosec(-x)sin(3/2pi-x))`
`=(-sinx(-sinx)cotx(-cotx))/(-sinxcosx-cosecx-cosx)`
`=(sin^2x*cos^2x)/(cos^2xsin^2x)`
`=1`.


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