Question

Prove that (tan2 theta-sin2 theta)=tan2 theta×sin2 theta

Answer 1

you mean,we have to prove (tan²θ - sin²θ) = tan²θ × sin²θ , right ?

LHS = tan²θ - sin²θ

= sin²θ/cos²θ - sin²θ

= sin²θ [ 1/cos²θ - 1]

[ we know, secx = 1/cosx so, 1/cos²θ = sec²θ]

= sin²θ [ sec²θ - 1]

we also know, sec²x - tan²x = 1

so, sec²θ - tan²θ = 1

or, sec²θ - 1 = tan²θ

then, sin²θ[sec²θ - 1] = sin²θ × tan²θ = RHS

hence, (tan²θ - sin²θ) = tan²θ × sin²θ

Answer 2

Step-by-step explanation:

The answer is attached you can check ✔it out ....!

Hope it will help!