[tex]\text{L.H.S}\\\\=(\tan q + \sec q)^2\\\\\\=\left(\dfrac{\sin q}{ \cos q} + \dfrac 1{ \cos q} \right)^2\\\\\\=\left( \dfrac{1+ \sin q}{\cos q}\right)^2\\\\\\=\dfrac{(1+ \sin q)^2}{\cos^2 q}\\\\\\=\dfrac{(1+ \sin q)^2}{1-\sin^2 q}\\\\\\=\dfrac{(1 + \sin q)(1 + \sin q)}{(1+ \sin q)(1 - \sin q)}\\\\\\=\dfrac{1+ \sin q}{1- \sin q}\\\\=\text{R.H.S}\\\\\text{Proved.}[/tex]
