If cosec θ - sin θ = a, sec θ - cos θ = b, prove that a ²b ²(a ²+b ²+ 3) = 1.

L.H.S = a ² b ² (a ² + b ² + 3)
= a ² b ² [(cosec θ - sin θ ) ² + (sec θ - cos θ ) ² + 3]
= a ² b ² [cosec ² θ + sin ² θ - 2 cosec θ sin θ + sec ² θ + cos ² θ - 2 sec θ cos θ + 3]
= a ² b ² [cosec ² θ + 1 - 2 + sec ² θ - 2 + 3]

= a ² b ² (cosec ² θ + sec ² θ)

Now, put the values of a and b again:

= (cosec θ - sin θ ) ² (sec θ - cos θ ) ² (cosec ² θ + sec ² θ)

= 1 = R.H.S.

Hence, proved.