1+cosθ
1+sinθ
×
1-cosθ
1-sinθ
= tan
2
θ
Answers
ANSWER
Given equations are
sinθ(1+sinθ)+cosθ(1+cosθ)=x
sinθ(1−sinθ)+cosθ(1−cosθ)=y
On solving both equations, we get
sinθ+sin 2
θ+cosθ+cos
2
θ=x
sinθ+cosθ+1=x→(1)
Similarly
sinθ(1−sinθ)+cosθ(1−cosθ)=y
sinθ−sin
2
θ+cosθ−cos
2
θ=y
sinθ+cosθ−1=y→(2)
On multiplying equation (1) and (2), we get
(sinθ+cosθ+1)(sinθ+cosθ−1)=xy
(sinθ+cosθ)
2
−1=xy
sin
2
θ+cos
2
θ+2sinθcosθ−1=xy
1+2sinθcosθ−1=xy
sin2θ=xy
Option C is correct
Similarly, we can check other options as
x
2
−2x−sin2θ=0
x
2
−2x=sin2θ
Taking LHS , we have
x
2
−2x
=(sinθ+cosθ+1)
2
−2(sinθ+cosθ+1)
=(sinθ+ cosθ+1)(sinθ+cosθ+1−2)
=(sinθ+cosθ)
2
−1
=sin
2
θ+cos
2
θ+2sinθcosθ−1
=1+2sinθcosθ−1
=sin2θ
=RHS
Similarly, we can prove other option
y
2
+2y−sin2θ=0
y
2
+2y=sin2θ
Taking LHS
=y
2
+2y
=y(y+2)
=(sinθ+cosθ−1)(sinθ+cosθ−1+2)
=(sinθ+cosθ)
2
−1
=sin
2
θ+cos
2
θ+2sinθcosθ−1
=1+2sinθcosθ−1
=sin2θ
=RHS