Question
If $\tan\theta=\frac{\text{a}}{\text{b}},$ prove that $\frac{\text{a}\sin\theta+\text{b}\cos\theta}{\text{a}\sin\theta+\text{b}\cos\theta}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}.$
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Answer
$\tan\theta=\frac{\text{a}}{\text{b}}.$ $\text{PT }\frac{\text{a}\sin\theta-\text{b}\cos\theta}{\text{a}\sin\theta+\text{b}\cos\theta}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}.$
Let $\frac{\text{a}\sin\theta-\text{b}\cos\theta}{\text{a}\sin\theta+\text{b}\cos\theta}\ \dots(1)$
Divide both Nr and Dr with $\cos\theta$ of (a)
$=\frac{\frac{\text{a}\sin\theta-\text{b}\cos\theta}{\cos\theta}}{\frac{\text{a}\sin\theta+\text{b}\cos\theta}{\cos\theta}}$
$=\frac{\text{a}\tan\theta-\text{b}}{\text{a}\tan+\text{b}}$
$=\frac{\text{a}\times\Big(\frac{\text{a}}{\text{b}}\Big)-\text{b}}{\text{a}\times\Big(\frac{\text{a}}{\text{b}}\Big)+\text{b}}$
$=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
Let $\frac{\text{a}\sin\theta-\text{b}\cos\theta}{\text{a}\sin\theta+\text{b}\cos\theta}\ \dots(1)$
Divide both Nr and Dr with $\cos\theta$ of (a)
$=\frac{\frac{\text{a}\sin\theta-\text{b}\cos\theta}{\cos\theta}}{\frac{\text{a}\sin\theta+\text{b}\cos\theta}{\cos\theta}}$
$=\frac{\text{a}\tan\theta-\text{b}}{\text{a}\tan+\text{b}}$
$=\frac{\text{a}\times\Big(\frac{\text{a}}{\text{b}}\Big)-\text{b}}{\text{a}\times\Big(\frac{\text{a}}{\text{b}}\Big)+\text{b}}$
$=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
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