please solve it..... it's urgent....
if sinθ + sin2θ = a and cosθ + cos2θ = b then prove that (a²+b²) (a²+b²-3) = 2b
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Step-by-step explanation:
Let
cosθ
1
=l and
sinθ
1
=m
The above expressions can be re-written as
ax(l)+by(m)=(a
2
−b
2
) ...(i)
ax(
m
l
2
)−by(
l
m
2
)=0 ..(ii)
Hence,
(axl)=
l
2
by(m
3
)
Substituting in (i) we get
by(m)(
l
2
m
2
+1)=(a
2
−b
2
)
bym(
l
2
m
2
+l
2
)=(a
2
−b
2
)
Now,
cosθ
1
=l and
sinθ
1
=m
Hence,
l
2
1
+
m
2
1
=1
l
2
+m
2
=l
2
m
2
Substituting we get.
by(m)(
l
2
l
2
m
2
)=(a
2
−b
2
)
by(m
3
)=(a
2
−b
2
)
sinθ=(
a
2
−b
2
by
)
1/3
...(iii)
Similarly we can get
cosθ=(
a
2
−b
2
ax
)
1/3
...(iv)
Squaring and adding (iii) and (iv) we get
(
a
2
−b
2
by
)
2/3
+(
a
2
−b
2
ax
)
2/3
=1
(by)
2/3
+(ax)
2/3
=(a
2
−b
2
)
2/3
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