# Question 696f9

Jun 19, 2016

Prove trig identity

#### Explanation:

Use these trig identities:
$1. \tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a . \tan b}$
$2. 1 + \cos 2 a = 2 {\cos}^{2} a$
3. sin 2a = 2sin a.cos a
Apply the first trig identity to the left side of the equation:
$L S = \tan \left(45 + \tan \left(\frac{t}{2}\right)\right) = \frac{1 + \tan \left(\frac{t}{2}\right)}{1 - \tan \left(\frac{t}{2}\right)} =$
Replace $\tan \left(\frac{t}{2}\right)$ by $\frac{\sin \frac{t}{2}}{\cos \frac{t}{2}}$, we get:
LS = (cos (t/2) + sin (t/2))/(cos (t/2) - sin (t/2)=  (1)
Now, transform the right side of the equation:
$R S = \frac{1 + \cos t + \sin t}{1 + \cos t - \sin t} =$
Replace
$1 + \cos t$ by $2 {\cos}^{2} \left(\frac{t}{2}\right)$
$\sin t = 2 \sin \left(\frac{t}{2}\right) . \cos \left(\frac{t}{2}\right)$, we get
RS = (2cos^2 (t/2)(cos (t/2) + sin (t/2)))/(2cos^2 (t/2)(cos (t/2) - sin (t/2)#.
$R S = \frac{\cos \left(\frac{t}{2}\right) + \sin \left(\frac{t}{2}\right)}{\cos \left(\frac{t}{2}\right) - \sin \left(\frac{t}{2}\right)} = L S$
The trig identity is proven.