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" If "([1,-tan theta],[tan theta,1])([1,...

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

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

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

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If [(1,-tan theta),(tan theta,1)][(1,tan theta),(-tan theta,1)]^(-1)=[(a,-b),(b,a)], then a) a = cos 2 θ , b = sin 2 θ b) a = 1 , b = 1 a = c) sin 2 θ , b = cos 2 θ d) a = sec 2 θ , b = tan 2 θ

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]]

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