If `E(theta)=[[cos theta, sin theta] , [-sin theta, cos theta]]` then `E(alpha) E(beta)=`

Options

A. `E(alpha beta)`
B. `E(alpha-beta)`
C. `E(alpha+beta)`
D. `-E(alpha+beta)`

Answer

C

Explanation

`E(theta)=[(cos theta,sin theta),(-sin theta,costheta)]`
Now `E(alpha)=[(cos alpha, sin alpha),(-sin alpha, cos alpha)]`
`and E(beta)=[(cos beta,sin beta),(-sin beta,cos beta)]`
Now, `E(alpha)E(beta)=[(cos alpha, sin alpha),(-sin alpha, cos alpha)][(cos beta sin beta),(-sin beta, cos beta)]`
`=[(cosalpha. cosbeta,cos alpha sin beta),(-sin alpha sin beta,+sin alpha cos beta),(-sin alpha cos beta,-sin alpha cos beta),(-cos alpha sin beta,+cos alpha cos beta)]`
`=[(cos (alpha+beta),sin(alpha+beta)),(-sin(alpha+beta),cos(alpha+beta))]`
`=E(alpha+beta)`
`therefore" Option (c) is correct."`