0²- 62 sino + 2ab coso=a²+b
Answers
Answer:
ANSWER
(a
2
−b
2
)sinθ+2abcosθ=a
2
+b
2
(dividingbycosθ)
= (a
2
−b
2
)
cosθ
sinθ
+2ab
cosθ
cosθ
=
cosθ
a
2
+b
2
= (a
2
−b
2
)tanθ+2ab=(a
2
+b
2
)secθ
= squaring both side we get
= (a
2
−b
2
)
2
tan
2
θ+4a
2
b
2
+4ab(a
2
−b
2
)tanθ=(a
2
+b
2
)
2
sec
2
θ
= (a
2
−b
2
)
2
tan
2
θ+4a
2
b
2
+4ab(a
2
−b
2
)tanθ=(a
2
+b
2
)
2
(1+tan
2
θ)
= [(a
2
−b
2
)
2
−(a
2
+b
2
)
2
]tan
2
θ+4a
2
b
2
+4ab(a
2
−b
2
)tanθ−(a
2
+b
2
)
2
=0
= [(a
2
−b
2
+a
2
+b
2
)(a
2
−b
2
−a
2
−b
2
)]tan
2
θ+4ab(a
2
−b
2
)tanθ+4a
2
b
2
−(a
4
+b
4
+2a
2
b
2
)=0
−4a
2
b
2
tan
2
θ+4ab(a
2
−b
2
)tanθ+2a
2
b
2
−a
4
−b
4
=0
+tan
2
θ−(
ab
a
2
−b
2
)tanθ−
2
1
+
4b
2
a
2
+
4a
2
b
2
=0
tan
2
θ+(
ab
b
2
−a
2
)tanθ−
2
1
+
4b
2
a
2
+
4a
2
b
2
=0
puttanθ=u,then
u
2
+(
ab
b
2
−a
2
)u−
2
1
+
4b
2
a
2
+
4a
2
b
2
=0
u=
2
−(
ab
b
2
−a
2
)±
(
ab
b
2
−a
2
)
2
−4(
4b
2
a
2
+
4a
2
b
2
−
2
1
)
=
2
ab
a
2
−b
2
±
4a
2
b
2
4(b
2
−a
2
)
2
−4(a
4
+b
4
)−2a
2
b
2
2
ab
a
2
−b
2
±
4a
2
b
2
4(b
2
−a
2
)
2
−4(a
4
+b
4
)−2a
2
b
2
=
2
ab
a
2
−b
2
±
4a
2
b
2
4b
4
+4a
4
−8a
2
b
2
−4a
4
−4b
4
+8a
2
b
2
tanθ=
2ab
a
2
−b
2
