Solve: `|[sintheta,tantheta],[-costheta,sintheta]|=2`

costheta[[costheta, sintheta], [-sintheta, costheta]]+sintheta[[sintheta, -costheta], [costheta, sintheta]]=

Evaluate- |[costheta,sintheta],[sintheta,costheta]|

costheta[{:(costheta,-sin theta),(sintheta,costheta):}]+sintheta[{:(sintheta,costheta),(-costheta,sintheta):}]=?

Consider a matrix A (theta)= [(sintheta,costheta),(-costheta,sintheta)] then

Evaluate Delta=|{:(sintheta, costheta),(-costheta,sintheta):}|

Show that costheta.[{:(costheta,sintheta),(-sintheta,costheta):}]+sintheta.[{:(sintheta,-costheta),(costheta,sintheta):}]=I.

Simplify: cos theta[{:(costheta,sintheta),(-sintheta,costheta):}]+sintheta[{:(sin theta ,-costheta),(costheta, sintheta):}]

Prove that : (sintheta-costheta)/(sintheta+costheta)+(sintheta+costheta)/(sintheta-costheta)=(2)/(2sin^(2)theta-1)

Show that every real orthogonal matrix is of any one of the forms {:[(costheta,-sintheta),(sintheta,costheta)]:}or{:[(costheta,sintheta),(sintheta,-costheta)]:}

Prove that : sintheta(1+tantheta)+costheta(1+cottheta)=cosectheta+sectheta