principal solution of c(acosB-bcosA) =
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MATHS
If sinA=acosB and cosA=bsinB then (a
2
−1)tan
2
A+(1−b
2
)tan
2
B is equal to (
b
2
a
2
−b
2
).
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ANSWER
Given:sinA=acosB and cosA=bsinB
On squaring and adding, we get
sin
2
A+cos
2
A=a
2
cos
2
B+b
2
sin
2
B
⇒1=a
2
cos
2
B+b
2
sin
2
B
⇒a
2
(1−sin
2
B)+b
2
sin
2
B=1
⇒(a
2
−b
2
)sin
2
B=a
2
−1
⇒sin
2
B=
a
2
−b
2
a
2
−1
⇒1=a
2
cos
2
B+b
2
sin
2
B
⇒1=a
2
cos
2
B+b
2
(1−cos
2
B)
⇒(a
2
−b
2
)cos
2
B=1−b
2
⇒cos
2
B=
a
2
−b
2
1−b
2
⇒tan
2
B=
cos
2
B
sin
2
B
=
a
2
−b
2
1−b
2
a
2
−b
2
a
2
−1
=
1−b
2
a
2
−1
and tan
2
A=
b
2
a
2
cot
2
B
=
b
2
a
2
.
a
2
−1
1−b
2
⇒(a
2
−1)tan
2
A+(1−b
2
)tan
2
B=
b
2
a
2
.(1−b
2
)+(a
2
−1)
⇒(a
2
−1)tan
2
A+(1−b
2
)tan
2
B=
b
2
a
2
−b
2