Question

sec theta minus sec squared theta + tan squared theta upon sec theta + sec squared theta minus tan squared theta is equal to 1 minus cos theta 1 + cos theta​

Answer 1

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