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

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

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

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]]

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

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

If "tan" 2 theta "tan" theta =1, "then" theta =

If "tan" 2 theta "tan" theta =1, "then" theta =

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