Show that `costheta` `[[costheta, sintheta],[-sintheta, cos theta]]`+`sin theta` `[[sintheta ,-cos theta],[costheta, sin theta]]`=1

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

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

If cos theta-4sintheta=1, the sintheta+4costheta=

(sintheta+sin2theta)/(1+costheta+cos2theta) =

(sintheta+sin2theta)/(1+costheta+cos2theta)=?

The value of f(pi/6) where f(theta)=|{:(cos^2theta,costhetasintheta,-sintheta),(costhetasintheta,sin^2theta,costheta),(sintheta,-costheta,0):}|

If sintheta - costheta = 0 , then sin^4theta - cos^4theta = ?

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

Evaluate the determinates abs([-costheta,-sin theta],[sintheta,-costheta])

inte^(sintheta)[log(sintheta)+cosec^(2)theta]costheta"d"theta=