Question
If $\tan\theta=\frac{\text{a}}{\text{b}},$ show that $\frac{(\text{a}\sin\theta-\text{b}\cos\theta)}{(\text{a}\sin\theta+\text{b}\cos\theta)}=\frac{\big(\text{a}^2-\text{b}^2\big)}{\big(\text{a}^2 +\text{b}^2\big)}.$
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Answer
Given:
$\tan\theta=\frac{\text{a}}{\text{b}}=\frac{\text{ak}}{\text{bk}}=\frac{\text{BC}}{\text{AB}}$
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theoram, we have
$\text{AC}^2=\text{AB}^2+\text{BC}^2=\text{b}^2\text{k}^2+\text{a}^2\text{k}^2$
$\therefore\text{AC}=\sqrt{\text{a}^2+\text{b}^2}\text{k}$
$\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{\text{ak}}{\sqrt{\text{a}^2+\text{b}^2}\text{k}}=\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2}}$
$\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{\text{bk}}{\sqrt{\text{a}^2+\text{b}^2}\text{k}}=\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2}}$
$\text{L.H.S.}=\frac{\text{a}\sin\theta-\text{b}\cos\theta}{\text{a}\sin\theta+\text{b}\cos\theta}$
$=\frac{\text{a}\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2}}-\text{b}.\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2}}}{\text{a}\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2}}+\text{b}.\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2}}}=\frac{\frac{\text{a}^2-\text{b}^2}{\sqrt{\text{a}^2+\text{b}^2}}}{\frac{\text{a}^2+\text{b}^2}{\sqrt{\text{a}^2+\text{b}^2}}}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
$=\text{R.H.S.}$
$\tan\theta=\frac{\text{a}}{\text{b}}=\frac{\text{ak}}{\text{bk}}=\frac{\text{BC}}{\text{AB}}$
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theoram, we have
$\text{AC}^2=\text{AB}^2+\text{BC}^2=\text{b}^2\text{k}^2+\text{a}^2\text{k}^2$
$\therefore\text{AC}=\sqrt{\text{a}^2+\text{b}^2}\text{k}$
$\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{\text{ak}}{\sqrt{\text{a}^2+\text{b}^2}\text{k}}=\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2}}$
$\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{\text{bk}}{\sqrt{\text{a}^2+\text{b}^2}\text{k}}=\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2}}$
$\text{L.H.S.}=\frac{\text{a}\sin\theta-\text{b}\cos\theta}{\text{a}\sin\theta+\text{b}\cos\theta}$
$=\frac{\text{a}\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2}}-\text{b}.\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2}}}{\text{a}\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2}}+\text{b}.\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2}}}=\frac{\frac{\text{a}^2-\text{b}^2}{\sqrt{\text{a}^2+\text{b}^2}}}{\frac{\text{a}^2+\text{b}^2}{\sqrt{\text{a}^2+\text{b}^2}}}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
$=\text{R.H.S.}$
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