Question

give me answer fast

Answer 1

According to given sum,


$= > \: \frac{ 1 - { \tan(45) }^{2} }{1 + { \tan(45) }^{2} }$

$= > \: \frac{1 - 1}{1 + 1}$

$= > \: \frac{0}{2}$

=> 0.


:-)Hope it helps u.

Answer 2

1 - tan^2 45° / 1 + tan^2 45°
= 1 - 1 / 1 + 1 [ tan^2 45° = 1]
= 0 / 2
= 0
= tan 0° [ tan^2 0° = 0]


Hope it helps u.... ☺️☺️