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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\frac{\sin(\theta+\phi)-2\sin\theta+\sin(\theta-\phi)}{\cos(\theta+\phi)-2\cos\theta+\cos(\theta-\phi)}=\tan\theta$
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Answer

We have,
$\text{LHS}=\frac{\sin(\theta+\phi)-2\sin\theta+\sin(\theta-\phi)}{\cos(\theta+\phi)-2\cos\theta+\cos(\theta-\phi)}$
$=\ \frac{\sin(\theta+\phi)+\sin(\theta-\phi)-2\sin\theta}{\cos(\theta+\phi)+\cos(\theta-\phi)-2\cos\theta}$
$=\ \frac{2\sin\Big[\frac{(\theta+\phi)+(\theta-\phi)}{2}\Big]\cos\Big[\frac{(\theta+\phi)-(\theta-\phi)}{2}\Big]-2\sin\theta}{2\cos\Big[\frac{(\theta+\phi)+(\theta-\phi)}{2}\Big]\cos\Big[\frac{(\theta+\phi)-(\theta-\phi)}{2}\Big]-2\cos\theta}$
$=\ \frac{2\sin(\theta)\cos(\phi)-2\sin(\theta)}{2\cos(\theta)\cos(\phi)-2\cos\theta}$
$=\ \frac{2\sin\theta(\cos\phi-1)}{2\cos\theta(\cos\phi-1)}$
$=\ \frac{\sin\theta}{\cos\theta}=\tan\theta$
$=\ \text{RHS}$
$\therefore\ \frac{\sin(\theta+\phi)-2\sin\theta+\sin(\theta-\phi)}{\cos(\theta+\phi)-2\cos\theta+\cos(\theta-\phi)}=\tan\theta$ Hence proved.

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