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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 38 of 61
Marks: +1, -0
cos⁡4x+cos⁡3x+cos⁡2xsin⁡4x+sin⁡3x+sin⁡2x\frac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x} = cot 3x
Solution:  
We have, L.H.S.= cos⁡4x+cos⁡3x+cos⁡2xsin⁡4x+sin⁡3x+sin⁡2x\frac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x}
=
2cos⁡(4x+2x2)cos⁡(4x−2x2)+cos⁡3x2sin⁡(4x+2x2)cos⁡(4x−2x2)+sin⁡3x\frac{2\cos\left(\frac{4x+2x}{2}\right)\cos\left(\frac{4x-2x}{2}\right)+\cos3x}{2\sin\left(\frac{4x+2x}{2}\right)\cos\left(\frac{4x-2x}{2}\right)+\sin3x}
= 2cos⁡3xcos⁡x+cos⁡3x2sin⁡3xcos⁡x+sin⁡3x\frac{2\cos3x\cos x+\cos3x}{2\sin3x\cos x+\sin3x} = cos⁡3x(2cos⁡x+1)sin⁡3x(2cos⁡x+1)\frac{\cos3x(2\cos x+1)}{\sin3x(2\cos x+1)} = cos⁡3xsin⁡3x\frac{\cos3x}{\sin3x} = cot 3x = R.H.S.
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