Prove please :)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Question

prove

sin^2y-sin^2x=1/(1+tan^2x)-1/(1+tan^2y)

Answer 1

Take R.H.S =1/(1+tan²x)-1/(1+tan²y)

Recall : Trigonometry formulas tan^2A = sin²A/cos²A

Substitute tan^2x = sin²x / cos²x in the R.H.S.

R.H.S =1/(1 + sin²x/cos²x) - 1(1+ sin²y / cos²y)

R.H.S =1/(cos²x +sin²x) / cos²x - 1/(cos²y + sin²y)/cos²y

Recall : Trigonometry forumalas sin²A +cos²A = 1

Substitute cos²x + sin²x = 1 and cos²y + sin²y = 1 in the R.H.S

R.H.S = 1/(1)/cos²x - 1/(1)/cos²y

R.H.S = cos²x - cos²y

Recall : Trigonometry formulas cos²A = 1 - sin²A

Substitute cos²x = 1 - sin²x and cos²y = 1 - sin²y

R.H.S = 1 - sin²x -(1 - sin²y)

Recall : Distributive property a(b - c) = ab - ac

R.H.S = 1 - sin²x -1 + sin²y

R.H.S = 0 + sin²y - sin²x

There fore R.H.S = sin²y - sin²x =  L.H.S

L.H.S. = R.H.S.

There fore sin^2y-sin^2x=1/(1+tan^2x)-1/(1+tan^2y)