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

0 votes

prove

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

asked Mar 25, 2013 in TRIGONOMETRY by M Gravity Rookie

1 Answer

+1 vote

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)

 

answered Mar 27, 2013 by diane Scholar

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