Trigonometry?

Question

Prove this identity:

tan^2(α)-sin^2(α)=sin^2(α)*tan^2(α)

*this is not tan^2α,this is tan(α)*tan(α)

Answer 1

tan²(a)-sin²(a)=sin²(a)*tan²(a)

(sin²a/cos²a) - sin²a = sin²a * tan²a

(sin²a - sin²a*cos²a)/cos²a = sin²a * tan²a

(sin²a)(1 - cos²a)/cos²a = sin²a * tan²a

using identity sin²x + cos²x = 1
and sin(x)/cos(x) = tan(x)

sin²a * tan²a = sin²a * tan²a
.

Answer 2

replace tan^2 (a) by sin^2 (a)/cos^2 (a) on left and then factor out sin^2 (a). Then you have
sin^2 (a) [sec^2 (a)-1] = sin^2 (a) * tan^2 (a)
I have a sneaky suspicision that the secant expresion equals tan^2 (a), which establishes the identity.

Answer 3

Identities
sin^2(α) + cos^2(α) = 1
sec^2(α) = 1 + tan^2(α)
sec^2(α) cos^2(α) = 1

RHS
=sin^2(α)*tan^2(α)
= (1 - cos^2(α)) (sec^2(α) - 1)
= sec^2(α) - sec^2(α) cos^2(α) - 1 + cos^2(α)
= sec^2(α) + cos^2(α) - 1 - 1
= sec^2(α) + cos^2(α) - 2
= 1 + tan^2(α) + 1 - sin^2(α) - 2
= tan^2(α) - sin^2(α)
= LHS