17. The expression
2
tan
𝜗
1−
tan2
𝜗
is identically equal to
A. cos 2𝜃
B. sin 2𝜃
C. tan 2𝜃
D. cot 2𝜃
Nope. C.
The horrible text formatting is hiding the identity
2tan𝜃/(1 - tan^2𝜃) = tan2𝜃
I apologize for the confusion. You are correct. The expression is equivalent to C. tan 2𝜃. Thank you for pointing out the error in the formatting.
To simplify the expression 2tanθ / (1 - tan^2θ), we can first apply the identity tan^2θ + 1 = sec^2θ.
2tanθ / (1 - tan^2θ)
= 2tanθ / [1 - (sec^2θ - 1)]
= 2tanθ / [2 - sec^2θ]
= 2tanθ / [2 - (1/cos^2θ)]
= 2tanθ / [(2cos^2θ - 1)/cos^2θ]
= 2tanθ * [cos^2θ/(2cos^2θ - 1)]
= 2sinθ/cosθ * [cos^2θ/(2cos^2θ - 1)]
= 2sinθ * [cos^2θ/(cosθ(2cos^2θ - 1))]
= 2sinθ * [cosθ/(2cos^2θ - 1)].
Now, we can use the double angle identity for sine, which states that sin2θ = 2sinθcosθ.
2sinθ * [cosθ/(2cos^2θ - 1)]
= [2sinθcosθ * cosθ/(2cos^2θ - 1)]
= sin2θ/(2cos^2θ - 1).
Therefore, the expression 2tanθ / (1 - tan^2θ) is identically equal to sin2θ/(2cos^2θ - 1), which is option B. sin 2θ.