If `A=[(0,-tan(alpha/2)),(tan(alpha/2),0)]`, then `I+A=(I-A)[(cos alpha,-sin alpha), (sin alpha, cos alpha)]`

A=[[0,-tan((alpha)/(2))tan((alpha)/(2)),0]]then(I-A)[[cos alpha,-sin alphasin alpha,cos alpha

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

If A=[(0,-tan (alpha/2)),(tan (alpha/2),0)] and I is 2 xx 2 unit matrix, then (I-A)[(cos alpha,-sin alpha),(sin alpha, cos alpha)] is

If A=[(0,-tan (alpha/2)),(tan (alpha/2),0)] and I is 2 xx 2 unit matrix, then (I-A)[(cos alpha,-sin alpha),(sin alpha, cos alpha)] is

If A=[(0,-tan (alpha/2)),(tan (alpha/2),0)] and I is 2 xx 2 unit matrix, then (I-A)[(cos alpha,-sin alpha),(sin alpha, cos alpha)] is

tan theta = (sin alpha-cos alpha) / (sin alpha + cos alpha)

If A=[(0,-tan alpha),(2,tan alpha),(2,0)] and I is 2 xx 2 unit matrix, then (I-A)[(cos alpha,sin alpha),(sin alpha, sin alpha)] is (a) -I+A (b) I-A (c) -I-A (d) non of these

If tan theta=(sin alpha- cos alpha)/(sin alpha+cos alpha) , then

If tan theta=(sin alpha- cos alpha)/(sin alpha+cos alpha) , then: