If asinx = bcosx = 2ctanx/(1-tan^2 x), then prove that (a^2 - b^2 )^2 = 4c^2 (a^2 + b^2 ).
Answers
Given:
To prove:
Taking square both sides
Apply Componendo and dividendo
Taking square both sides
Cross multiply
hence proved
Answer:
Given: a\sin x=b\cos x=\dfrac{2c\tan x}{1-\tan^2x}asinx=bcosx=
1−tan
2
x
2ctanx
To prove: (a^2-b^2)^2=4c^2(a^2+b^2)^2(a
2
−b
2
)
2
=4c
2
(a
2
+b
2
)
2
a\sin x=b\cos xasinx=bcosx
\dfrac{a}{b}=\dfrac{\cos x}{\sin x}
b
a
=
sinx
cosx
Taking square both sides
\dfrac{a^2}{b^2}=\dfrac{\cos^2 x}{\sin^2 x}
b
2
a
2
=
sin
2
x
cos
2
x
Apply Componendo and dividendo
\dfrac{a^2+b^2}{a^2-b^2}=\dfrac{\cos^2x+\sin^2 x}{\cos^2x-\sin^2 x}
a
2
−b
2
a
2
+b
2
=
cos
2
x−sin
2
x
cos
2
x+sin
2
x
\dfrac{a^2+b^2}{a^2-b^2}=\dfrac{1}{\cos2x}
a
2
−b
2
a
2
+b
2
=
cos2x
1
\because \cos^2x+\sin^2 x=1\text{ and }\cos^2x-\sin^2 x=\cos 2x∵cos
2
x+sin
2
x=1 and cos
2
x−sin
2
x=cos2x
Taking square both sides
\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\right)^{2}=\left(\frac{1}{\cos2x}\right)^{2}(
a
2
−b
2
a
2
+b
2
)
2
=(
cos2x
1
)
2
\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\right)^{2}=\left(\frac{\sec^4x\cdot b^2\cos^2x}{4c^2\tan^2x}\right)(
a
2
−b
2
a
2
+b
2
)
2
=(
4c
2
tan
2
x
sec
4
x⋅b
2
cos
2
x
)
\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\right)^{2}=\dfrac{a^2+b^2}{4c^2}(
a
2
−b
2
a
2
+b
2
)
2
=
4c
2
a
2
+b
2
Cross multiply
(a^2-b^2)^2=4c^2(a^2+b^2)(a
2
−b
2
)
2
=4c
2
(a
2
+b
2
)
hence proved
Step-by-step explanation:
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