# Verify if the following is true: 1/cos^2x+tan^2y = 1/cos^2y+tan^2x

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### 2 Answers

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.**

First, we'll substitute 1/(cos x)^2 = (sec x)^2 and 1/(cos y)^2 = (sec y)^2

We'll re-write the identity:

(tan x)^2 - (tan y)^2 = (sec x)^2 - (sec y)^2

We notice the differences of squares from both sides:

(tan x)^2 - (tan y)^2 = (tan x - tan y)(tan x + tan y)

(sec x)^2 - (sec y)^2 = (sec x - sec y)(sec x + sec y)

sec x - sec y = 1/cos x - 1/cos y

sec x - sec y = (cos y - cos x)/cos y*cos x

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

sec x + sec y = 1/cos x + 1/cos y

sec x + sec y = (cos y + cos x)/cos y*cos x

cos y + cos x = 2cos[(x+y)/2]cos[(x-y)/2]

(sec x - sec y)(sec x + sec y) = sin 2[(x+y)/2]*sin 2[(x-y)/2]/(cos y*cos x)^2

We'll simplify and we'll get:

**(sec x - sec y)(sec x + sec y) = sin (x+y)*sin (x-y)/(cos y*cos x)^2**

We'll work now on the left side of the equal:

tan x = sin x/cos x

tan y = sin y/cos y

tan x - tan y = (sin x*cosy - siny*cosx)/cos x*cos y

tan x - tan y = sin (x-y)/cos x*cos y

tan x + tan y = sin (x+y)/cos x*cos y

**(tan x - tan y)(tan x + tan y) = sin (x-y)*sin (x-y)/(cos x*cos y)^2**

**Identity is verified since we've get the same expression, both sides of the equal sign.**