If `[(1,-tan theta), (tan theta, 1)][(1,tan theta), (-tan theta,1)]^(-1) = [(a,-b), (b,a)],` then

([1, -tan thetatan theta, 1]) ([1, tan theta-tan theta, 1]) (- 1) = [[a, -bb, a]]

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

Show that [[1 , -tan theta ],[ tan theta, 1]][[1, tan theta],[ -tan theta, 1]]^(-1)=[[cos 2 theta, -sin 2 theta],[ sin 2 theta, cos 2 theta]]

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

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

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

If tan theta + tan 2theta + tan theta tan2theta =1 then theta =

If tan2 theta*tan theta=1, then theta=