# Prove this identity: tan x = (sin (x + 2pi))/(cos(x - 2pi))

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tan(x) = sin(x+2pi) / cos(x-2pi)

We will use trigonometric identities to prove the equality.

We know that:

sin(a+ b) = sina*cosb+ cosa*sinb

==> sin(x+2pi) = sinx*cos2pi + cosx*sin2pi

==> sin(x+2pi) = sinx*1 + cosx* 0 = sinx,..........(1)

Now we know that:

cos(a-b) = cosa*cosb- sina*sinb

==> cos(x-2pi) = cosx*cos2pi + sinx*sin2pi

= cosx*1 + sinx*0 = cosx.............(2)

Now, from (1) and (2) we conclude that:

sin(x+2pi)/cos(x-2pi) = sinx/cos x = tanx .........q.e.d

**==> Then we prove that tanx = sin(x+2pi)/cos(x-2pi)**

The sine and cosine functions are periodic with a periodicity equal to 2*pi. For any angle x, sin x = sin (x + n*2*pi) and cos x = (cos x + n*2*pi) with n being any integer.

sin (x + 2*pi) = sin x

cos (x - 2*pi) = cos x

(sin (x + 2*pi)) /cos (x - 2*pi)

=> sin x / cos x

=> tan x

**This proves the given identity.**