Question

prove that cos(a+b)=b^2-a^2/b^2+a^2

Answer 1

What is the question?

Answer 2

Um let a be pi/2 and b be pi/2 then cos (a+b) is -1

Answer 3

But the rhs is 0

Answer 4

Maybe we should assume \(a\) and \(b\) are distinct?

Answer 5

This is not true

Answer 6

Same thing but use a is -pi/2 and u get 1 on lbs and 0 on rhs

Answer 7

Lhs*

Answer 8

It could be, then, that we have to assume \(|a|\not=|b|\), or that the asker meant the rhs to be
\[b^2-\frac{a^2}{b^2}+a^2\]