Question

If sec theta -tan theta = a +1/a - 1
then cos theta is equal to​

Answer 1

$\textbf{Given:}$

$\sec\theta-\tan\theta=\displaystyle\frac{a+1}{a-1}$........(1)

$\text{We know that}$

$\boxed{\bf\,sec^2\theta-\tan^2\theta=1}$

$\text{Using}$

$\boxed{a^2-b^2=(a-b)(a+b)}$

$\implies(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$

$\implies\displaystyle\frac{a+1}{a-1}(\sec\theta+\tan\theta)=1$

$\implies\sec\theta+\tan\theta=\displaystyle\frac{a-1}{a+1}$........(2)

$\text{Adding (1) and (2), we get}$

$2\sec\theta=\displaystyle\frac{a+1}{a-1}+\frac{a-1}{a+1}$

$2\sec\theta=\displaystyle\frac{(a+1)^2+(a-1)^2}{(a-1)(a+1)}$

$2\sec\theta=\displaystyle\frac{a^2+1+2a+a^2+1-2a}{a^2-1}$

$2\sec\theta=\displaystyle\frac{2a^2+2}{a^2-1}$

$2\sec\theta=\displaystyle\frac{2(a^2+1)}{a^2-1}$

$\text{Cancelling 2 on both sides,}$

$\sec\theta=\displaystyle\frac{a^2+1}{a^2-1}$

$\text{Taking reciprocals on both sides, we get}$

$\implies\boxed{\bf\:\cos\theta=\frac{a^2-1}{a^2+1}}$