It says, prove:

2csc^2(2A) -csc^2(A) = -2cot(2A)csc(2A)

where A = theta

As usual, develop the LHS to see if it matches the RHS.

LHS =

= 2csc^2(2A) -csc^2(A)

= [2 /sin^2(2A)] -[1 /sin^2(A)]

= [2 / (2sinAcosA)^2] -[1 /sin^2(A)]

= [1 / 2sin^2(A)cos^2(A)] -[1 /sin^2(A)]

Combine the two fractions into one fraction only,

= [1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]

= [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]

= [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

= [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]

= [-cot(2A)]*[2 /sin(2A)]

= [-cot(2A)]*[2csc(2A)]

= -2cot(2A)csc(2A)

= RHS

Proven.