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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove that:
$\frac{(\sin7\text{x}+\sin5\text{x})+(\sin9\text{x}+\sin3\text{x})}{(\cos7\text{x}+\cos5\text{x})(\cos9\text{x}+\cos3\text{x})}=\tan6\text{x}$
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Answer

$\text{L.H.S.}=\frac{(\sin7\text{x}+\sin5\text{x})+(\sin9\text{x}+\sin3\text{x})}{(\cos7\text{x}+\cos5\text{x})(\cos9\text{x}+\cos3\text{x})}$$=\frac{\Big[2\sin\Big(\frac{7\text{x}+5\text{x}}{2}\Big)\cos\Big(\frac{7\text{x}-5\text{x}}{2}\Big)\Big]+\Big[2\sin\Big(\frac{9\text{x}+3\text{x}}{2}\Big)\cos\Big(\frac{9\text{x}-3\text{x}}{2}\Big)\Big]}{\Big[2\cos\Big(\frac{7\text{x}+5\text{x}}{2}\Big)\cos\Big(\frac{7\text{x}-5\text{x}}{2}\Big)\Big]+\Big[2\cos\Big(\frac{9\text{x}+3\text{x}}{2}\Big)\cos\Big(\frac{9\text{x}-3\text{x}}{2}\Big)\Big]}$
$=\frac{2\sin6\text{x}\cos\text{x}+2\sin6\text{x}\cos3\text{x}}{2\cos6\text{x}\cos\text{x}+2\cos6\text{x}\cos3\text{x}}$
$=\frac{2\sin6\text{x}(\cos\text{x}+\cos3\text{x})}{2\cos6\text{x}(\cos\text{x}+\cos3\text{x})}$
$=\tan6\text{x}=\text{R.H.S.}$

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