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Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles5 Marks
Question
Prove that:
$\frac{1}{\sin\text{(x}-\text{b})\sin\text{(x}-\text{b)}}=\frac{\cot\text{(x}-\text{b)}-\cot\text{(x}-\text{b)}}{\sin\text{(a}-\text{b)}}$
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Answer

$\text{L.H.S}\ \frac{1}{\sin\text{(x}-\text{a)}\sin\text{(x}-\text{b)}}$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(a}-\text{b)}}{\sin\text{(x}-\text{b})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(x}-\text{b)-}(\text{x}-\text{b)}}{\sin\text{(x}-\text{a})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(x}-\text{b)}\cos(\text{x}-\text{a})-\cos(\text{x}-\text{b})\sin(\text{x}-\text{a)}}{\sin\text{(x}-\text{a})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin(\text{x}-\text{b)}\cos(\text{x}-\text{a})}{\sin\text{(x}-\text{a)}\sin(\text{x}-\text{b})}-\frac{\sin(\text{x}-\text{b)}\cos(\text{x}-\text{a})}{\sin\text{(x}-\text{a)}\sin(\text{x}-\text{b})}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\cot(\text {x}-\text{a}-\cot(\text{x}-\text{b})\Big]$
$=\frac{\cot(\text{x}-\text{a})-\cot(\text{x}-\text{b})}{\sin(\text{a}-\text{b})}$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.

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