Please help using identities
(2tan30)/(1-tan^2 30)
[sin/cos] / [ 1 - sin^2/cos^2 ]
sin cos / [cos^2-sin^2]
sqrt3/4 / [ 3/4 - 1/4 ]
(1/2) sqrt 3
Remember your double-angle formulas. You have
tan(2*30) = √3
Reiny dropped the 2 in the numerator...
oops, sorry Reiny...
To simplify the expression (2tan30)/(1-tan^2 30), we can use the identity for tangent:
tan^2 θ + 1 = sec^2 θ
Let's start by looking at the denominator, (1 - tan^2 30):
Since we know that tan 30 = 1/√3, we can substitute this value:
(1 - (1/√3)^2) = (1 - 1/3) = (3/3 - 1/3) = 2/3
Now, let's simplify the numerator, 2tan 30:
2tan 30 = 2 * (1/√3)
To simplify this further, we can rationalize the denominator by multiplying the numerator and denominator by √3:
(2 * (1/√3)) * (√3/√3) = (2√3)/(√3 * √3) = (2√3)/(√(3^2)) = (2√3)/(√9) = (2√3)/3
Now, after simplifying the numerator and denominator, we have:
(2√3)/3 / 2/3 = (2√3)/3 * 3/2
Now, cancel out the common factors in the numerator and denominator:
(2√3 * 3)/(3 * 2) = (6√3)/(6)
Finally, we have the simplified expression:
(6√3)/(6) = √3
So, the simplified expression is √3.