Can someone solve this identity pls

(tan²θ - sin²θ)/sin²θ = tan²θ

(tan²(θ) - sin²(θ))/sin²(θ)
= csc²(θ)(tan²(θ) - sin²(θ))
= sec²(θ) - 1
= tan²θ

Don't listen to that idiot - here is a better way:

(tan²θ - sin²θ)/sin²θ = tan²θ (multiply by sin²θ)

tan²θ - sin²θ = sin²θtan²θ (factorise sin²θ)

tan²θ = sin²θ(1 + tan²θ) (1+tan²θ = sec²θ)

tan²θ/sec²θ = sin²θ

sin²θ = sin²θ


θ = whatever the **** you want it to be.

Given:
(tan² θ − sin² θ)/sin² θ = tan² θ

Note that tan² θ = (sin² θ)/(cos² θ):
((sin² θ)/(cos² θ) − sin² θ)/sin² θ = tan² θ

Find a common denominator for the terms in the numerator:
[(sin² θ)/(cos² θ) − (sin² θ)(cos² θ)/(cos² θ)]/sin² θ = tan² θ

Combine:
[((sin² θ) − (sin² θ)(cos² θ))/(cos² θ)]/sin² θ = tan² θ

Factor out sin² θ:
[sin² θ(1 − cos² θ)/(cos² θ)]/sin² θ = tan² θ

Use the identity 1 − cos² θ = sin² θ:
[(sin^4 θ)/(cos² θ)]/sin² θ = tan² θ

Divide by the sin² θ in the denominator:
(sin² θ)/(cos² θ) = tan² θ

And thus:
tan² θ = tan² θ

QED

(tan²θ - sin²θ)/sin²θ
= tan²θ/sin²θ - 1
= (sin²θ/cos²θ)/sin²θ - 1
= 1/cos²θ - 1
= sec²θ - 1
= tan²θ