Vectors Q Cambridge (1 Viewer)

[hmim]

Can someone please explain why in the solution to this question, there is a restriction of y can't equal to 1 please?

 

[unofficiallyred12]

View attachment 38643
View attachment 38644
Can someone please explain why in the solution to this question, there is a restriction of y can't equal to 1 please?

(1-t^2)/(1+t^2) = (-1-t^2+2)/(1+t^2) = -1 + 2/(1+t^2) since 2/(1+t^2) is always positive, y can't equal -1

 

[hmim]

(1-t^2)/(1+t^2) = (-1-t^2+2)/(1+t^2) = -1 + 2/(1+t^2) since 2/(1+t^2) is always positive, y can't equal -1

If it's always positive, why didn't they mention that y cant equal to all negative values, but only mention y can't equal to -1?

 

[unofficiallyred12]

2/(1+t^2) is always positive not y

 

[hmim]

2/(1+t^2) is always positive not y

Ohhhh right since 2/1+t^2 can never = 0