Cos ^ 2A * Tan ^ 2A = sin ^ 2A? Why

Cos ^ 2A * Tan ^ 2A = sin ^ 2A? Why

tana=(sina)/(cosa)
Then:
tan²a=(sin²a)/(cos²a_
Namely:
coa²atan²a=sin²a

If a is the first quadrant angle, how many of sin (2a + π), cos (- 2A), Sina / 5, cosa / 5, Tan (A / 2 - π) can be determined as the fixed value?
A 1, B.2, C.3, D.4

sin(2a+π),tan(a/2-π)

Given Tan (a + Pai / 4) = 2, find 1 + 2sinacosa / cos ^ 2a-sin ^ 2A

tan(a+π/4)=2=(tana+tanπ/4)/(1-tanatanπ/4)=(tana+1)/(1-tana)
tana=1/3
(1+2sinacosa)/(cos^2a-sin^2a)
=(sin ^ 2A + cos ^ 2A + 2sinacosa) / (COS ^ 2a-sin ^ 2a) (divide cos ^ 2a)
=((tan^2a+1+2tana)/(1-tan^2a)
=(1/9+1+2/3)/(1-1/9)
=2