Question

tan (π/4+x) + tan (π/4-x) = 2sec2x .

Answer 1

tan(pi/4 + x) + tan(pi/4 - x) = 
(1+tan x)/(1-tan x)+ (1- tan x)/(1+tan x) = 
[(1+tan x)^2 +(1-tan x)^2]/[(1-tan^2 x] = 
2(1+tan^2 x)/(1-tan^2 x) = 
2(cos^2 x+sin^2 x)/(cos^2x-sin^2 x) = 
2/cos2x = 2sec2x 

Answer 2

Answer:

Step-by-step explanation:

The given equation is:

$tan({\frac{\pi}{4}+x)+tan({\frac{\pi}{4}-x)=2sec2x$

Taking the LHS of the above equation, we get

$tan({\frac{\pi}{4}+x)+tan({\frac{\pi}{4}-x)=\frac{1+tanx}{1-tanx}+\frac{1-tanx}{1+tanx}$

=$\frac{(1+tanx)^{2}+(1-tanx)^2}{1-tan^2x}$

=$\frac{2(1+tan^2x)}{(1-tan^2x)}$

=$\frac{2(cos^2x+sin^2x)}{cos^2x-sin^2x}$

=$\frac{2}{cos2x}$

=$2sec2x$

=RHS

Hence proved