if sin A=(a²–b²)/(a²+b²) find the valu of all t–ratio
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We have,
sin A =
Hypotenuse
Perpendicular
=
a
2
+b
2
a
2
−b
2
So, we draw a right triangle right angled at B such that
Perpendicular = a
2
−b
2
and, Hypotenuse = a
2
+b
2
. and ∠BAC=θ
By Pythagoras theorem, we have
AC
2
=AB
2
+BC
2
⇒AB
2
=(a
2
+b
2
)
2
−(a
2
−b
2
)
2
⇒AB
2
=(a
4
+b
4
+2a
2
b
2
)−(a
4
+b
4
−2a
2
b
2
)
⇒AB
2
=4a
2
b
2
=(2ab)
2
⇒AB=2ab
When we consider the trigonometric ratios of ∠BAC=θ , we have
Base = AB = 2ab, Perpendicular = BC = a
2
−b
2
, and Hypotenuse = AC = a
2
+b
2
∴cosθ=
Hypotenuse
Base
=
a
2
+b
2
2ab
⇒tanθ=
Base
Perpendicular
=
2ab
a
2
−b
2
⇒cosecθ=
Perpendicular
Hypotenuse
=
a
2
−b
2
a
2
+b
2
⇒secθ=
Base
Hypotenuse
=
2ab
a
2
+b
2
and,
cotθ=
Perpendicular
Base
=
a
2
−b
2
2ab