We have to prove that 1/cos^2x+tan^2y=1/cos^2y+tan^2x

Now 1/( cos x)^2 + ( tan y )^2 - 1/ (cos y)^2 - (tan x)^2

=> [(sin y)^2/ (cos y)^2] + 1/ (cos x)^2 - 1/ (cos y)^2 - [(sin x)^2/ (cos x)^2]

=> [1 - (sin x)^2]/ (cos x)^2 - [ ( 1 - (sin y)^2]/ (cos y )^2

we know that (sin x)^2 + (cos x)^2 = 1 or 1 - (sin x)^2 = (cos x)^2

=> (cos x)^2 / (cos x)^2 - (cos x)^2 / (cos x)^2

=> 1 - 1

=> 0

So 1/( cos x)^2 + ( tan y )^2 - 1/ (cos y)^2 - (tan x)^2 = 0

=> 1/( cos x)^2 + ( tan y )^2 = 1/ (cos y)^2 + (tan x)^2

**Therefore 1/( cos x)^2 + ( tan y )^2 = 1/ (cos y)^2 + (tan x)^2.**