In other words....
we have to prove that
2sinxcosx + 4cos^2 x - 1 = tan^2 x - 2tanx - 3
I had my doubts about this "identity" so I tried any arbitrary value of x
(remember , all we need is ONE exception and the identity is not true)
I tried x = 21°
LS = 2sin21 cos21 + 4cos^2 21° - 1 = appr 3.155..
RS = tan^2 21° - 2tan21 - 3 = appr -3.62
So it cannot be proven
Check your typing.
prove that the equation 2sin x cos x +4cos^2 x =1 may be written in the form of tan^2 x -2tan x -3=0
4 answers
the two functions are clearly not the same, since one is pure sines and cosines (just a bunch of wavy stuff), and the other is pure tangents (full of asymptotes).
Firstly split the 4cos^2X into cos^2x + 3cos^2x.
Then 1-cos^2x will be sin^2x
The eq then will be 2sinxcosx+3cos^2x=sin^2x
Next divide the eq with cos^2x then u will get the ans
Then 1-cos^2x will be sin^2x
The eq then will be 2sinxcosx+3cos^2x=sin^2x
Next divide the eq with cos^2x then u will get the ans
divide both sides of the eq by cos^2x -> 2tanx+4=1/cos^2x=(sin^2x+cos^2x)/cos^2x=tan^2x+1 so tan^2x-2tanx-3=0