Answer 1

We need to prove cosec⁶θ = cot⁶θ + 3 cot²θ cosec²θ + 1

Solving the L.H.S, we get

`cosec^6 theta = (cosec^2 theta)^3`

`= (1 + cot^2 theta)^3`     .......`(1 + cot^2 theta = cosec^2 theta)`

Further using the identity `(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2`  we get

`(1 + cot^2 theta)^3 = 1 + cot^6 theta + 3(1)^2 (cot^2 theta) + 3(1) (cot^2 theta)^2`

`= 1 + cot^6 theta + 3 cot^2 theta + 3 cot^4 theta`

`= 1 + cot^6 theta + 3 cot^2 theta (1 + cot^2 theta)`

`= 1 + cot^6 theta + 3 cot^2 theta cosec^2 theta`    `(using 1 + cot^2 theta = cosec^2 theta)`

Hence proved.