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costheta[(costheta, sintheta),(-sintheta...

`costheta[(costheta, sintheta),(-sintheta,costheta)]+sintheta[(sintheta,-costheta),(costheta,sintheta)]` is equal to (A) `[(0,0),(1,1)]` (B) `[(1,1),(0,0)]` (C) `[(0,1),(1,0)]` (D) `[(1,0),(0,1)]`

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Show that: |(sintheta,costheta),(-costheta,sintheta)| =1

(1+sintheta-costheta)/(1+sintheta+costheta) =

What is sintheta/(1+costheta)+(1+costheta)/sintheta equal to ?

(1+sintheta-costheta)/(1+sintheta+costheta)=

(1+costheta+sintheta)/(1+costheta-sintheta)=(1+sintheta)/(costheta)

Prove: (costheta)/(1+sintheta)=(1-sintheta)/(costheta)

If sintheta-costheta=1 ,then find sintheta.costheta

(1+costheta+sintheta)/(1+costheta-sintheta)=(1+sintheta)/costheta .

If A=[(costheta,-sintheta),(sintheta,costheta)] " then " A^(-1) =?