(cos (pi+x).cos(-x))/(sin(pi-x).cos(pi/2...

`(cos (pi+x).cos(-x))/(sin(pi-x).cos(pi/2+x))` = `cot^2x`

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Given,
L.H.S
`[cos(pi+x)cos(-x)]/[ sin(pi-x)cos(pi/2+x)]`
`=[ -(cosxcosx)]/[-(sinxsinx)]`
...

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