Question

Cos theta = (a^2+b^2/2ab) where a and b are two distinct numbers such that ab>0. is it true

Answer 1

Answer:

FALSE

Step-by-step explanation:

We now that (a-b)²≥0

=> a² + b² -2ab≥0

=>a² + b² ≥ 2ab

=>(a² + b²)/2ab ≥1

But cos theta is always ≤ 1.

Suppose cos theta = 1

=> (a² + b²)/2ab = 1

=> (a² + b²) = 2ab

=> (a-b)² = 0

=> a = b

But given that a an b are distinct number,

hence cos theta > 1.

But we now that range of cos theta is [-1,1].

Hence , false

cos theta cannot take the value of (a^2+b^2/2ab).

Answer 2

Answer:

the above is the answer

Step-by-step explanation:

hope it helps