How do you verify the identity $tan (\frac{A}{2}) = \frac{sin A}{1 + cos A}$ $tan(A/2)=sinA/(1+cosA)$ ?

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Answer 1 · 2016-12-14T17:43:42

See verification below

We use

$sin 2 θ = 2 sin θ cos θ$ $sin2theta=2sintheta costheta$

and $cos 2 θ = 2 {cos}^{2} θ - 1$ $cos2theta=2cos^2theta-1$

We deduce,

$sin A = 2 sin (\frac{A}{2}) cos (\frac{A}{2})$ $sinA=2sin(A/2)cos(A/2)$

and, $cos A = 2 {cos}^{2} (\frac{A}{2}) - 1$ $cosA=2cos^2(A/2)-1$

$:. R.H.S=sinA/(1+cosA)=(2sin(A/2)cos(A/2))/(1+2cos^2(A/2)-1)$

$=(2sin(A/2)cos(A/2))/(2cos^2(A/2))=sin(A/2)/cos(A/2)$

$=tan(A/2)=LHS$

Q. E.D

How do you verify the identity tan(A/2)=sinA/(1+cosA)tan(A2)=sinA1+cosAtan(A/2)=sinA/(1+cosA)?