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Prove that if sin theta (1+ sin theta) + cos theta (1+ cos theta) = x and sin theta (1- sin theta) + cos theta (1- cos theta) = y then x^2 - 2x - sin 2 theta = y^2 + 2y - sin 2 theta = 0

Prove that if sin theta (1+ sin theta) + cos theta (1+ cos theta) = x
and sin theta (1- sin theta) + cos theta (1- cos theta) = y
then x^2 - 2x - sin 2 theta = y^2 + 2y - sin 2 theta = 0

Grade:11

1 Answers

Aditya Gupta
2081 Points
5 years ago
hi ananya, thanks for asking.
let sin(theta)=s and cos(theta)=c
x= s+s^2+c+c^2=s+c+1
or 2x=2s+2c+2
and x^2=s^2+c^2+1+2(sc+c+s)=2+2(sc+s+c)
so, x^2 - 2x - sin 2 theta = 2sc – sin 2 theta =0 [becoz  sin 2 theta =2sin theta* cos theta]
similarly, y=s+c – 1
so, 2y=2s+2c – 2
and y^2=2+2(sc-s-c)
so, y^+2y= 2sc
or y^2 + 2y - sin 2 theta = 0

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