If `f(x)=[cosx-sinx -sin x cosc 1] and g(y)=[cosy sin y siny cos y],` then `[f(x)g(y)]^-1` is equal to (a) `f(-x)g(-y)` (b) `g(-y)f(-x)` (c) `f(x^-1)g(y^-1)` (d) `g(y^-1)f(x^-1)`

F(x)=[[cosx,-sinx,0],[sinx,cosx,0],[0,0,1]] and G(x)=[[cosx,0,sinx],[0,1,0],[-sinx,0,cosx]], then [F(x)G(y)]^(-1) is equal to (A) F(-x)G(-y) (B) F(x-1)G(y-1) (C) G(-y)F(-x) (D) G(y^(-1))F(x^(-1))

If F(x)=[("cos"x,-sin x,0),(sin x,cos x,0),(0,0,1)] and G(y)=[(cos y,0,sin y),(0,1,0),(-sin y,0,cos y)] , then [F(x) G(y)]^(-1) is equal to

Let f(x)=[(cosx,-sinx,0),(sinx,cosx,0),(0,0,1)], then (A) (f(x))^2=-I (B) f(x+y)=f(x),f(y) (C) f(x)^-1=f(-x) (D) f(x)^-1=f(x)

If f(x) = [(cos x , - sinx,0),(sinx,cosx,0),(0,0,1)] then show f(x) . f(y) = f(x+y)

f(x)=[(cosx,-sinx,0),(sinx,cosx,0),(0,0,1)] Statement 1: f(x) is inverse of f(-x) Statement 2: f(x).f(y) = f(x+y)

Given the graph of y=f(x) . Draw the graphs of the followin. (a) y=f(1-x) (b) y=-2f(x) (c) y=f(2x) (d) y=1-f(x)