`( sec theta + tan theta -1)/( tan theta - sec theta +1)`

(tan theta+sec theta-1)/(tan theta-sec theta+1)

The inverse of [(sec theta, - tan theta),(-tan theta, sec theta)] is

prove that (tan theta) / (sec theta -1) = (tan theta + sec theta + 1) / (tan theta + sec theta - 1)

Simplify: tan theta[(sec theta, tan theta),(tan theta, -sec theta)] + sec theta[(- tan theta, -sec theta),(-sec theta, tan theta)] .

Prove that: (tantheta)/(sectheta -1) = (tan theta + sec theta + 1)/(tan theta + sec theta -1)

Prove : (sec theta-1)/(tan theta)=(tan theta)/(sec theta+1)

( tan theta)/( sec theta -1)-(tan theta )/( sec theta + 1)=

Prove that : (sec theta-tan theta)/(sec theta+tan theta)= 1-2 sec theta tan theta+ 2 tan^2 theta .