If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal to

A(alpha, beta)=((cos alpha, sin alpha, 0),(- sin alpha, cos alpha, 0),(0,0,e^(beta)))=[A(alpha, beta)]^(-1)=

If f(alpha,beta)=|(cos alpha,-sin alpha,1),(sin alpha,cos alpha,1),(cos(alpha+beta),-sin(alpha+beta),1)|, then

If f(alpha,beta)=|(cos alpha,-sin alpha,1),(sin alpha,cos alpha,1),(cos(alpha+beta),-sin(alpha+beta),1)|, then

If F(alpha)=[[cos alpha,-sin alpha, 0],[sin alpha, cos alpha, 0],[0,0,1]] , then show that F(alpha) F (beta)= F(alpha + beta) .

The value of the determinant ,cos alpha,-sin alpha,1sin alpha,cos alpha,1cos(alpha+beta),-sin(alpha+beta),1]| is equal

IF cos ( alpha + beta ) =0 then sin ( alpha + 2 beta ) is equal to

Find the value of f(alpha, beta) =|(cos (alpha + beta), -sin (alpha + beta), cos 2beta),(sin alpha, cos alpha, sin beta),(-cos alpha, sin alpha, cos beta)|

If cos alpha+cos beta=0=sin alpha+sin beta, then cos2 alpha+cos2 beta is equal to

Delta=det[[0,sin alpha,-cos alpha-sin alpha,0,sin betacos alpha,-sin beta,0]]