Answer:
sin^2α
Step-by-step explanation:
I will just pose x instead of alpha here to make things simpler
[tex]\tan ^2\left(x\right)\left(2\cos ^2\left(x\right)+\sin ^2\left(x\right)-1\right)\\\\[/tex]
we know that sin^2x = 1 - cos^2x so...
[tex]=\left(-1+1-\cos ^2\left(x\right)+2\cos ^2\left(x\right)\right)\tan ^2\left(x\right)[/tex]
[tex]=\cos ^2\left(x\right)\tan ^2\left(x\right)[/tex]
we can rewrite using trigonometric identities (tan = sin/cos)...
[tex]=\left(\frac{\sin \left(x\right)}{\cos \left(x\right)}\right)^2\cos ^2\left(x\right)\\= \sin ^2\left(x\right)[/tex]
