Question

if tan theta = a/b prove that {(a sin theta) -(b cos theta )}/{(a sin theta) +(b cos theta )}=( a²-b²)/(a²+b²)

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Answer 1

I give you 2 methods ..take a look

Answer 2

Answer:  The proof is done below.

Step-by-step explanation:  We are given the following :

$\tan\theta=\dfrac{a}{b}.$

We are to prove the following relation :

$\dfrac{a\sin\theta-b\cos\theta}{a\sin\theta+b\cos\theta}=\dfrac{a^2-b^2}{a^2+b^2}.$

We will be using the following trigonometric formula :

$\dfrac{\sin A}{\cos A}=\tan A.$

We have

$L.H.S.\\\\\\=\dfrac{a\sin\theta-b\cos\theta}{a\sin\theta+b\cos\theta}\\\\\\=\dfrac{\dfrac{a\sin\theta-b\cos\theta}{\cos\theta}}{\dfrac{a\sin\theta+b\cos\theta}{\cos\theta}}~~~~~~~~~~~~~~[\textup{dividing both nemerator and denominator by }\cos\theta]\\\\\\=\dfrac{a\dfrac{\sin\theta}{\cos\theta}-b}{a\dfrac{\sin\theta}{\cos\theta}+b}\\\\\\=\dfrac{a\tan\theta-b}{a\tan\theta+b}\\\\\\=\dfrac{a\times\dfrac{a}{b}-b}{a\times\dfrac{a}{b}+b}\\\\\\=\dfrac{\dfrac{a^2-b^2}{b}}{\dfrac{a^2+b^2}{b}}\\\\\\=\dfrac{a^2-b^2}{a^2+b^2}\\\\=R.H.S.$

Hence proved.