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3.Trigonometrical Ratios, Functions and Identities
medium
निम्नलिखित को सिद्ध कीजिए
$\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$
Option A
Option B
Option C
Option D
Solution
$L.H.S.$ $=\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}$
$=\frac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x}$
$=\frac{2 \cos \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+\cos 3 x}{2 \sin \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+\sin 3 x}$
$[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right),$
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)]$
$=\frac{2 \cos 3 x \cos +\cos 3 x}{2 \sin 3 x \cos x+\sin 3 x}$
$=\frac{\cos 3 x(2 \cos x+1)}{\sin 3 x(2 \cos x+1)}$
$\cot 3 x=R .H .S.$
Standard 11
Mathematics
