Which double-angle or half-angle identity would you use to verify that 1+cos2a=2/1+tan2a?

In this case or scenario, the double-angle identity that should be used is the one for cosine. 

In totality, we shall need the following three trigonometric identities to end up with the equality: 

1. cos (2a) = cos² (a) - sin² (a)
2. sin² (a) + cos² (a) = 1 
3.  tan² (a) + 1 = sec² (a)

Using identities 1 and 2 on the left-hand side of the equation, we get the following:

1 + cos (2a) = 1 + cos² (a) - sin² (a) = 2 cos² (a)

Recalling that cos² (a) = 1 / sec² (a) and applying identity 3, we find the following:

2 cos² (a) = 2 / sec² (a) = 2 / (1 + tan² (a))

Therefore giving us:

2 cos² (a) = 2 / (1 + tan² (a))

b

step-by-step explanation:

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct