Question

Prove that : $\frac{tan\theta }{sec\theta -1}$ - $\frac{tan\theta }{sec\theta +1}$ = 2 cot$\theta$

Answer 1

LHS = $\frac{tan\theta }{sec\theta -1}$ - $\frac{tan\theta }{sec\theta +1}$

= $\frac{tan\theta (sec\theta +1)-tan\theta (sec\theta -1)}{se{c}^2\theta -1}$

= $\frac{tan\theta sec\theta +tan\theta -tan\theta sec\theta -tan\theta }{se{c}^2\theta -1}$

= $\frac{2tan\theta }{ta{n}^2\theta }$

= $\frac{2}{tan\theta }=2cot\theta$ (RHS)