Question

Simplify (1-cos theta)(1+cos theta) over (1-sin theta) (1+ sin theta)

Answer 1

`\bf sin^2(\theta)+cos^2(\theta)=1\to \begin{cases} cos^2(\theta)=1-sin^2(\theta)\\ sin^2(\theta)=1-cos^2(\theta) \end{cases} \\\\ -------------------------------\\\\ \stackrel{\textit{difference of squares}}{\cfrac{[1-cos(\theta )][1+cos(\theta )]}{[1-sin(\theta )][1+sin(\theta )]}}\implies \cfrac{1^2-cos^2(\theta )}{1^2-sin^2(\theta )}\implies \cfrac{1-cos^2(\theta )}{1-sin^2(\theta )} \\\\\\ \cfrac{sin^2(\theta )}{cos^2(\theta )}\implies tan^2(\theta )`