Question

If tan theta equal to a by b then prove that a sin theta minus cot theta by a sin theta + b cos theta equal to a square minus b square by a square + b square

Answer 1

Given:

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

To Prove:

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

Solution:

$\tan \theta = \frac{a}{b} \\ \\ or \: \: \frac{ \sin \theta}{ \cos \theta} = \frac{a}{b} \\ \\ or \: \: \frac{a}{b} \times \frac{ \sin \theta}{ \cos \theta} = \frac{a}{b} \times \frac{a}{b} \\ \\ \frac{ a\sin \theta}{ b\cos \theta} = \frac{ {a}^{2} }{ {b}^{2}} \\ \\ [\text{Using Dividendo and Componendo}]\\ \\ \dfrac{a \sin\theta-b\cos\theta}{a \sin\theta + b\cos\theta} = \dfrac{{a}^{2}-{b}^{2}}{ {a}^{2}+{b}^{2}} \\ \\ \therefore \text{L.H.S = R.H.S [Proved]}$

Important Rules

$\sin \theta = \frac{1}{ \cosec} \\ \\ \cos = \frac{1}{ \sec} \\ \tan \theta \: = \frac{1}{ \cot \theta} = \frac{ \sin\theta}{ \cos \theta}$

Componendo:

$\frac{a}{b} = \frac{c}{d} \\ \\ \frac{a + b}{b} = \frac{c + d}{d} \\ \\ or \: \: \frac{a}{b + a} = \frac{c}{d + c}$

Dividendo:

$\frac{a}{b} = \frac{c}{d} \\ \\ \frac{a-b}{b} = \frac{c - d}{d} \\ \\ or \: \: \frac{a}{b - a} = \frac{c}{d - c}$

Answer 2

Answer:

$\bf{Given :}\:\sf\tan(\theta)=\dfrac{a}{b}$

$\bf{To\:Prove :}\:\sf\dfrac{a\sin(\theta) - b\cos(\theta)}{a\sin(\theta) + b\cos(\theta)}=\dfrac{a^2 - b^2}{a^2 + b^2}$

$\rule{120}{1}$

$\bf{Proof :}$

$\dashrightarrow\sf\:\:\tan(\theta)=\dfrac{a}{b} = \dfrac{\sin(\theta)}{\cos(\theta)}\\\\\\\dashrightarrow\sf\:\:a = \sin(\theta) \quad and \quad b = \cos(\theta)$

$\underline{\bigstar\:\textbf{According to the Question :}}$

$:\implies\sf\dfrac{a\sin(\theta) - b\cos(\theta)}{a\sin(\theta) + b\cos(\theta)}\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Putting the values.}}\\\\:\implies\sf \dfrac{a(a) - b(b)}{a(a) + b(b)}\\\\\\:\implies\sf \underline{\boxed{\sf \dfrac{a^2 - b^2}{a^2 + b^2}}}\\\\\\\therefore\:\sf\dfrac{a\sin(\theta) - b\cos(\theta)}{a\sin(\theta) + b\cos(\theta)}=\dfrac{a^2 - b^2}{a^2 + b^2}\\\\{\:\qquad\underline{\bf{\dag}\:\mathcal{HENCE, PROVED}}}$