# sec^2 x - sec^2 y = tan^2 x - tan^2 y

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...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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.

Approved by eNotes Editorial Team