Question

its a true or false and ans is false
but how pls explain.......​

Answer 1

Answer:

it's false

Step-by-step explanation:

cos0 = 1/sec0

Answer 2

Answer:

FALSE

Step-by-step explanation:

$\: as \:(a - b )^{2} \geqslant 0$

=> a² + b² -2ab≥0

=>a² + b² ≥ 2ab

=>(a² + b²)/2ab ≥1

But, cos theta is always ≤ 1.

$suppose \: cosθ = 1$

=> (a² + b²)/2ab = 1

=> (a² + b²) = 2ab

=> (a-b)² = 0

=> a = b

But given that a an b are distinct number,

hence cosθ> 1.

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

Hence, False

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

i hope it helps you ^_^