Answer:
Step-by-step explanation:
[tex]1+tan^2\theta=sec^2\theta\\tan^2\theta=sec^2\theta -1 \\tan^2\theta=(sec\theta-1)(sec\theta+1)\\\implies \frac{1}{sec\theta-1} =\frac{sec\theta+1}{tan^2\theta} \\\\and \quad \frac{1}{sec\theta+1} =\frac{sec\theta-1}{tan^2\theta} \\\\ \frac{1}{sec\theta-1} + \frac{1}{sec\theta+1} = \frac{sec\theta+1}{tan^2\theta} + \frac{sec\theta-1}{tan^2\theta}\\\\=\dfrac{2sec\theta}{tan^2\theta}\\\\=\dfrac{2cos^2\theta}{cos\theta.sin^2\theta}\\\\\\=2cos\theta.cosec^2\theta[/tex]
