Question

(sin^2Q/cosQ+1)-(sin^2Q/cosQ-1)

Answer 1

((2sinQcosQ / cosQ) +1 ) - ((2 sinQcosQ / cost ) - 1)
= 2sinQ + 1 - 2sinQ + 1 = 2

Answer 2

Hey there !!!!!

=sin²θ/cosθ+1 -sin²θ/cosθ-1-----Equation 1

We know that sin²θ+cos²θ =1

So, sin²θ=1-cos²θ and cos²θ=1-sin²θ

Substituting these in equation 1

=(sin²θ/cosθ+1) -(sin²θ/cosθ-1)

=(1-cos²θ/1+cosθ)-(1-cos²θ/cosθ-1)---- Equation 2

Now using a²-b²=(a+b)(a-b) in equation 2

=(1+cosθ)(1-cosθ)/1+cosθ - (1-cosθ)(1+cosθ)/-(1-cosθ)

=1-cosθ-(-1+cosθ)

=1-cosθ+1-cosθ

=2

Hope this helped you...................