What are two expressions that are equivalent to tan(θ)sin(θ)?

Answer 1

Since tan(θ) = sin(θ)/cos(θ),

tan(θ)sin(θ) = (sin(θ)/cos(θ)) * sin(θ) = sin²(θ)/cos(θ)

Furthermore, since sin²(θ) = 1 - cos²(θ), we can substitute into our previous answer such that

tan(θ)sin(θ) = (1 - cos²(θ))/cos(θ) = 1/cos(θ) - cos²(θ)/cos(θ) = sec(θ) - cos(θ)

Thus,

tan(θ)sin(θ) = sin²(θ)/cos(θ)

tan(θ)sin(θ) = sec(θ) - cos(θ)