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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 57 of 61
Marks: +1, -0
(sin⁡7x+sin⁡5x)+(sin⁡9x+sin⁡3x)(cos⁡7x+cos⁡5x)+(cos⁡9x+cos⁡3x)\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}
Solution:  
We have,
L.H.S. = (sin⁡7x+sin⁡5x)+(sin⁡9x+sin⁡3x)(cos⁡7x+cos⁡5x)+(cos⁡9x+cos⁡3x)\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}
2sin⁡(7x+5x2)cos⁡(7x−5x2)2\sin\left(\frac{7x+5x}{2}\right)\cos\left(\frac{7x-5x}{2}\right)
= +2sin⁡(9x+3x2)cos⁡(9x−3x2)2cos⁡(7x+5x2)cos⁡(7x−5x2)\frac{+2\sin\left(\frac{9x+3x}{2}\right)\cos\left(\frac{9x-3x}{2}\right)}{2\cos\left(\frac{7x+5x}{2}\right)\cos\left(\frac{7x-5x}{2}\right)}
+ 2cos⁡(9x+3x2)cos⁡(9x−3x2)2\cos\left(\frac{9x+3x}{2}\right) \cos\left(\frac{9x-3x}{2}\right)
Since cos A + cos B = 2 cos (A+B2)cos⁡(A−B2)\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)
sin A + sin B = 2 sin (A+B2)cos⁡(A−B2)\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)
= 2sin⁡6xcos⁡x+2sin⁡6xcos⁡3x2cos⁡6xcos⁡x+2cos⁡6xcos⁡3x\frac{2\sin 6x \cos x + 2\sin 6x \cos 3x}{2\cos 6x \cos x + 2\cos 6x \cos 3x} = 2[cos⁡x+cos⁡3x]sin⁡6x2[cos⁡x+cos⁡3x]cos⁡6x]\frac{2[\cos x + \cos 3x] \sin 6x}{2[\cos x + \cos 3x] \cos 6x]}
= sin⁡6xcos⁡6x\frac{\sin 6x}{\cos 6x} = tan 6x = R.H.S.
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