Difference of Squares in Trigonometric Functions
$$\frac{ 3tanx }{ 1+\tan^2 }=3sinxcosx$$

Question

Difference of Squares in Trigonometric Functions
$$\frac{ 3tanx }{ 1+\tan^2 }=3sinxcosx$$

anonymous · Answer

Are you trying to prove that the LHS = RHS ?

anonymous · Answer

Yes

anonymous · Answer

ok.

anonymous · Answer

what are you stuck with? Do you understand how to approach this problem?

anonymous · Answer

I mean, I understand that you need to factor in order to solve it but I don't know where to start

anonymous · Answer

$$(3tanx)/(1+\tan^2x) = 3sinxcosx$$
$$Since,  tanx = sinx/cosx$$
We can do the following:
$$3sinx/cosx/(1+ \sin^2x/\cos^2x)$$

anonymous · Answer

then find a common denominator which is cos^x

anonymous · Answer

$$3sinx/cosx * \cos^2x / \cos^2x+\sin^2x$$
Then we notice that cos^2 + sinx^2 = 1 <- which is the Pythagorean identity.

anonymous · Answer

then after more algebraic simplification we get 3sinxcosx

anonymous · Answer

Ahh thank you!

anonymous · Answer

thus both sides match and the identity holds.

anonymous · Answer

Your welcome.