# How do you verify the identity  tan(x/2 + pi/4) = secx + tanx?

Mar 9, 2018

See Below

#### Explanation:

We will use the following Properties:

color(blue)(sin(x+y)=sinxcosy+cosxsiny

color(blue)(cos(x+y)=cosxcosy-sinxsiny

color(blue)(sin2x=2sinxcosx

color(blue)(cos2x=cos^2x-sin^2x

color(blue)(sin^2x+cos^2x=1

$L H S : \tan \left(\frac{x}{2} + \frac{\pi}{4}\right)$

=sin(x/2+pi/4)/(cos(x/2+pi/4)

$= \frac{\sin \left(\frac{x}{2}\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(\frac{x}{2}\right) \sin \left(\frac{\pi}{4}\right)}{\cos \left(\frac{x}{2}\right) \cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{x}{2}\right) \sin \left(\frac{\pi}{4}\right)}$

=(1/sqrt2 sin(x/2)+1/sqrt2 cos (x/2))/(1/sqrt2cos(x/2)-1/sqrt2sin(x/2)

=(1/sqrt2( sin(x/2)+ cos (x/2)))/(1/sqrt2(cos(x/2)-sin(x/2))->factor out common factor

=(cancel(1/sqrt2)( sin(x/2)+ cos (x/2)))/(cancel(1/sqrt2)(cos(x/2)-sin(x/2))-> cancel common factor

$= \frac{\sin \left(\frac{x}{2}\right) + \cos \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right)}$

$= \frac{\sin \left(\frac{x}{2}\right) + \cos \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right)} \cdot \frac{\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)}$->multiply by conjugate

$= \frac{{\sin}^{2} \left(\frac{x}{2}\right) + 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) + {\cos}^{2} \left(\frac{x}{2}\right)}{{\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right)}$

$= \frac{\left[{\sin}^{2} \left(\frac{x}{2}\right) + {\cos}^{2} \left(\frac{x}{2}\right)\right] + 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{{\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right)}$

$= \frac{1 + 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{{\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right)}$

$= \frac{1 + \sin 2 \left(\frac{x}{2}\right)}{\cos 2 \left(\frac{x}{2}\right)}$

$= \frac{1 + \sin \cancel{2} \left(\frac{x}{\cancel{2}}\right)}{\cos \cancel{2} \left(\frac{x}{\cancel{2}}\right)}$

$= \frac{1 + \sin x}{\cos} x$

$= \frac{1}{\cos} x + \sin \frac{x}{\cos} x$

$= \sec x + \tan x$

$= R H S$