If `A = [[0,-tan alpha/2],[tan alpha/2,0]]` and I is the identity matrix of order 2, show that `I + A = (I - A)[[cos alpha,-sin alpha],[sin alpha,cos alpha]]`

Explanation

Since, `I=[(1,0),(0,1)]` and given `A = [(0,-tan(alpha//2)),(tan(alpha//2),0)]`
`therefore" " 1+A = [(1,-tan(alpha//2)),(tan(alpha//2),1)]`
`RHS = (I-A)[(cosalpha,-sinalpha),(sinalpha,cosalpha)]`
`=[(1,tan(alpha//2)),(-tan(alpha//2),1)][(cosalpha,-sinalpha),(sinalpha,cosalpha)]`
`=[(1,tan(alpha//2)),(-tan(alpha//2),1)]`
`[((1-tan^(2)alpha//2)/(1tan^(2)(alpha//2)),(-2tan(alpha//2))/(1+tan^(2)(alpha//2))),((2tan(alpha//2))/(1+tan^(2)(alpha//2)),(1-tan^(2)(alpha//2))/(1+tan^(2)(alpha//2)))]`
`RHS = [(1,lambda),(-lambda,1)][((1-lambda^(2))/(1+lambda^(2)),(-2lambda)/(1+lambda^(2))),((2lambda)/(1+lambda^(2)),(1-lambda^(2))/(1+lambda^(2)))]`
`[((1-lambda^(2)+2lambda^(2))/(1+lambda^(2))(-2lambda+lambda(1-lambda^(2)))/(1+lambda^(2))),((-lambda(1-lambda)^(2)+2lambda)/(l+lambda^(2))(2lambda^(2)+1-lambda^(2))/(1+lambda^(2)))]`
`[((1+lambda^(2))/(1+lambda^(2))( lambda(1+lambda)^(2))/(1+lambda^(2))),((lambda(1+lambda^(2)))/(1+lambda^(2))(1+lambda^(2))/(1+lambda^(2)))]=[(1,-lambda),(lambda,1)]`
`=[(1,-tan(alpha//2)),(-tan(alpha//2),1)][therefore lambda =(alpha//2)]`
`=I+A`
`=LHS`