Question #2bbce

May 19, 2016

Assuming $\cos \left(x\right) \ne 0$ and $\sin \left(x\right) \ne 0$, using the identity ${\csc}^{2} \left(x\right) = 1 + {\cot}^{2} \left(x\right)$, we have

${\csc}^{2} \left(x\right) {\tan}^{2} \left(x\right) - 1 = \left(1 + {\cot}^{2} \left(x\right)\right) {\tan}^{2} \left(x\right) - 1$

$= {\tan}^{2} \left(x\right) + {\cot}^{2} \left(x\right) {\tan}^{2} \left(x\right) - 1$

$= {\tan}^{2} \left(x\right) + {\left(\cot \left(x\right) \tan \left(x\right)\right)}^{2} - 1$

$= {\tan}^{2} \left(x\right) + {1}^{2} - 1$

$= {\tan}^{2} \left(x\right)$