if tan θ = a/b , show that a sin θ - b cos θ/ a sin θ + b cos θ = a² - b²/a²+b²
Answers
Step-by-step explanation:
Step-by-step explanation: We are given the following :
\tan\theta=\dfrac{a}{b}.tanθ=
b
a
.
We are to prove the following relation :
\dfrac{a\sin\theta-b\cos\theta}{a\sin\theta+b\cos\theta}=\dfrac{a^2-b^2}{a^2+b^2}.
asinθ+bcosθ
asinθ−bcosθ
=
a
2
+b
2
a
2
−b
2
.
We will be using the following trigonometric formula :
\dfrac{\sin A}{\cos A}=\tan A.
cosA
sinA
=tanA.
We have
\begin{gathered}L.H.S.\\\\\\=\dfrac{a\sin\theta-b\cos\theta}{a\sin\theta+b\cos\theta}\\\\\\=\dfrac{\dfrac{a\sin\theta-b\cos\theta}{\cos\theta}}{\dfrac{a\sin\theta+b\cos\theta}{\cos\theta}}~~~~~~~~~~~~~~[\textup{dividing both nemerator and denominator by }\cos\theta]\\\\\\=\dfrac{a\dfrac{\sin\theta}{\cos\theta}-b}{a\dfrac{\sin\theta}{\cos\theta}+b}\\\\\\=\dfrac{a\tan\theta-b}{a\tan\theta+b}\\\\\\=\dfrac{a\times\dfrac{a}{b}-b}{a\times\dfrac{a}{b}+b}\\\\\\=\dfrac{\dfrac{a^2-b^2}{b}}{\dfrac{a^2+b^2}{b}}\\\\\\=\dfrac{a^2-b^2}{a^2+b^2}\\\\=R.H.S.\end{gathered}
L.H.S.
=
asinθ+bcosθ
asinθ−bcosθ
=
cosθ
asinθ+bcosθ
cosθ
asinθ−bcosθ
[dividing both nemerator and denominator by cosθ]
=
a
cosθ
sinθ
+b
a
cosθ
sinθ
−b
=
atanθ+b
atanθ−b
=
a×
b
a
+b
a×
b
a
−b
=
b
a
2
+b
2
b
a
2
−b
2
=
a
2
+b
2
a
2
−b
2
=R.H.S.
Hence proved.