::

To prove :

(Sin^4(θ) – cos^4(θ) + 1)(cosec^2(θ) =2

Left hand side :

(Sin^4(θ) – cos^4(θ)+1)(cosec^2(θ))

=> We can write sin^4(θ) as (sin^2(θ))^2

and cos^4(θ) as (cos^2(θ))^2 .

=> ((Sin^2(θ))^2 -(cos^2(θ))^2 +1)(cosec^2(θ))

=> Now we will use the following identity: (a^2 – b^2) = (a+b)(a-b) in our equation

=> ((Sin^2(θ)-cos^2(θ))(sin^2(θ) + cos^2(θ) ) + 1)(cosec^2(θ))

=> We know that :

(sin^2(θ) + cos^2(θ))= 1 ,

Therefore,

((Sin^2(θ)-cos^2(θ)).1 +1)(cosec^2(θ))

=> We can write it as ↓

=> (Sin^2(θ) +1- cos^2(θ))(cosec^2(θ))

=> We can write sin^2(θ) as

(1- cos^2(θ)) as we know sin^2(θ)+cos^2(θ) = 1

sin^2(θ) = 1- cos^2(θ) —{eq a}

=> (sin^2(θ) + sin^2(θ))(cosec^2(θ)). {using eq a}

=> (2sin^2(θ))(cosec^2(θ))

=> Using the relation between cosec(θ) and sin(θ)

We know sin(θ)=1/(cosec(θ) or

Cosec(θ)=1/(sin(θ)

Squaring both side

cosec^2(θ)=1/(sin^2(θ)). —eq 1

=> (2sin^2(θ))/(sin^2(θ)). {using eq 1}

=> 2

Hence proved