Question

if x+y=π\4 than find the value of (1+tan x)(1+tan y)​

Answer 1

Answer:

Given,

x + y = \frac{\pi}{4}x+y=4π

= > tan(x + y) = tan \frac{\pi}{4}=>tan(x+y)=tan4π

\frac{tanx + tany}{1 - tanx.tany} = 11−tanx.tanytanx+tany=1

tan x + tan y = 1 - tan x tan y

tan x + tan y + tan x tan y = 1

Adding 1 on both sides

1 + tan x + tan y + tan x tan y = 2

(1+tan x) + tan y (1+tan x)

☆_________________________

==> ( 1+tan x) ( 1+ tan y) = 2

_________________________☆