We start with the more complicated side which is the left side, and show that, on using some trigonometric identities, we will get the term on the right side .
[tex]\frac{sin \theta + tan \theta}{1+cos \theta}[/tex]
Using Quotient identity for tangent function, we will get
[tex]\frac{sin \theta+ \frac{sin \theta}{cos \theta}}{1+cos \theta}[/tex]
[tex]\frac{sin \theta cos \theta + sin \theta}{cos \theta(1+cos \theta)}[/tex]
Taking out sine function from the numerator
[tex]=\frac{sin \theta(1+cos \theta)}{cos \theta(1+cos \theta)}[/tex]
Cancelling the common term of numerator and denominator
[tex]=\frac{sin \theta}{cos \theta} = tan \theta[/tex]
