Show that `AB=BA`where, `A=[[costheta,sintheta],[-sintheta,costheta]],B=[[costheta,-sintheta],[sintheta,costheta]]`

Show that AB=BA if A=[[costheta,sintheta],[-sintheta,costheta]],B=[[costheta,-sintheta],[sintheta,costheta]]

Answer the following questions.If A=[[costheta,sintheta],[-sintheta,costheta]] ,prove that A^n=[[cos ntheta,sin ntheta],[-sin ntheta,cos ntheta]] , for all n in N

Simplify costheta[[costheta,sintheta],[-sintheta,costheta]+sintheta[[sintheta,-costheta],[costheta,sintheta]]

(sintheta+sin 2theta)/(1+costheta+cos2theta)=

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

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

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

Solve the following: Fing theta ,if [[sintheta],[costheta],[theta]][sintheta,costheta,theta]=[17]

If costheta-sintheta=sqrt2sintheta then costheta+sintheta is equal to

Prove that (frac(sintheta-2sin^3theta)(2cos^3theta-costheta))=tantheta