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Trigonometric Ratios of Multiple and Submultiple Angles — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios of Multiple and Submultiple Angles4 Marks
Question
Prove that: $\frac{1}{\cos\text{(x}-\text{b})\cos\text{(x}-\text{b)}}=\frac{\tan\text{(x}-\text{a)}-\tan\text{(x}-\text{b)}}{\sin\text{(a}-\text{b)}}$
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Answer

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

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