Prove that:
(a) `(cos(pi+x) cos(-x))/(sin(pi-x)cos(pi/2+x))=cot^(2)x`
(b) `cos((3pi)/2 + x)cos(2pi+x){cot ((3pi)/2-x)+cot(2pi+x)}=1`

Explanation

(a) We have,
LHS `=(cos(pi+x)cos(-x))/(sin(pi-x)cos(pi/2+x))`
`=-(cos^(2)x)/(-sin^(2)x)=cot^(2)x` =RHS
Hence, proved
(b) We have,
LHS `=cos((3pi)/2+x) cos(2pi+x){(cot(3pi)/2-x) + cot(2pi+x)}`
`=(sinx) (cosx)(tanx+cotx) [therefore cos((3pi)/2+x)=sin,cos(2pi+x)=cosx, cot((3pi)/2-x)=tanx]` and `cot(2pi+x)=cotx]`
`=sinx cosx[(sinx)/(cosx)+(cosx)/(sinx)]=sinx cosx{(sin^(2)x+cos^(2)x)/(sinx cosx)}`
`=(sinx cosx) xx 1/(sinx cosx)=1` =RHS
Hence Proved.