We have to prove that (sec x)^2 - (sec y )^2 = (tan x)^2 - (tan y)^2.

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

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

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

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

=> (sec x)^2 - (sec y )^2 = 1/(cos x)^2 - (cos x)^2/(cos x)^2 - 1/(cos y)^2- (cos y)^2/(cos y)^2

=> (sec x)^2 - (sec y )^2 = (sec x)^2 - 1 - (sec y)^2 + 1

=> (sec x)^2 - (sec y )^2 = (sec x)^2 - (sec y)^2 + 1- 1

=> **(sec x)^2 - (sec y )^2 = (sec x)^2 - (sec y)^2 **

Therefore the two sides are equal. Hence the relation is proved.

L:H:S ≡ sec²x - sec²y

**⇒ use the identity sec²θ = 1 + tan²θ **

= 1 + tan²x -(1 + tan²y)

= 1 + tan²x -1 - tan²y

= tan²x - tan²y

= R:H:S

We notice the differences of squares from both sides:

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

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

We also know that sec x = 1/cos x and sec y = 1/cos 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 right 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**

We notice that we've get the same expression both sides, so the identity is verified.

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