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Miscellaneous Exercise · Q6

Q.Prove that (sin⁡7x+sin⁡5x)+(sin⁡9x+sin⁡3x)(cos⁡7x+cos⁡5x)+(cos⁡9x+cos⁡3x)=tan⁡6x\dfrac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x.

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The expression simplifies to tan⁡6x\tan 6x by grouping sine and cosine sums into products using sum-to-product identities, then cancelling the common factor 2cos⁡x2 \cos x.

We need to prove that the given fraction equals tan⁡6x\tan 6x. The numerator and denominator each contain four trigonometric terms. The direct approach — expanding each term — would be messy. Instead, notice that the angles inside each pair of sines (and cosines) are symmetric around 6x6x: 7x7x and 5x5x average to 6x6x, and 9x9x and 3x3x also average to 6x6x. This is a classic cue to use the sum-to-product identities, which convert a sum of two sines or two cosines into a product involving the average angle and half the difference.

The sum-to-product formulas you need are:

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2 \sin\frac{A+B}{2} \cos\frac{A-B}{2}

cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2 \cos\frac{A+B}{2} \cos\frac{A-B}{2}

These identities let us rewrite the numerator and denominator as products, and then a common factor will cancel, leaving a single tangent.

Let’s work through it step by step.

  1. Group the sine terms in the numerator. Pair (sin⁡7x+sin⁡5x)(\sin 7x + \sin 5x) and (sin⁡9x+sin⁡3x)(\sin 9x + \sin 3x) separately. For the first pair: sin⁡7x+sin⁡5x=2sin⁡7x+5x2cos⁡7x−5x2=2sin⁡6xcos⁡x\sin 7x + \sin 5x = 2 \sin\frac{7x+5x}{2} \cos\frac{7x-5x}{2} = 2 \sin 6x \cos x. For the second pair: sin⁡9x+sin⁡3x=2sin⁡9x+3x2cos⁡9x−3x2=2sin⁡6xcos⁡3x\sin 9x + \sin 3x = 2 \sin\frac{9x+3x}{2} \cos\frac{9x-3x}{2} = 2 \sin 6x \cos 3x. So the numerator becomes:

2sin⁡6xcos⁡x+2sin⁡6xcos⁡3x=2sin⁡6x (cos⁡x+cos⁡3x).2 \sin 6x \cos x + 2 \sin 6x \cos 3x = 2 \sin 6x \, (\cos x + \cos 3x).

  1. Group the cosine terms in the denominator similarly. Pair (cos⁡7x+cos⁡5x)(\cos 7x + \cos 5x) and (cos⁡9x+cos⁡3x)(\cos 9x + \cos 3x). For the first pair: cos⁡7x+cos⁡5x=2cos⁡7x+5x2cos⁡7x−5x2=2cos⁡6xcos⁡x\cos 7x + \cos 5x = 2 \cos\frac{7x+5x}{2} \cos\frac{7x-5x}{2} = 2 \cos 6x \cos x. For the second pair: cos⁡9x+cos⁡3x=2cos⁡9x+3x2cos⁡9x−3x2=2cos⁡6xcos⁡3x\cos 9x + \cos 3x = 2 \cos\frac{9x+3x}{2} \cos\frac{9x-3x}{2} = 2 \cos 6x \cos 3x. So the denominator becomes:

2cos⁡6xcos⁡x+2cos⁡6xcos⁡3x=2cos⁡6x (cos⁡x+cos⁡3x).2 \cos 6x \cos x + 2 \cos 6x \cos 3x = 2 \cos 6x \, (\cos x + \cos 3x).

  1. Now the whole fraction is: 2sin⁡6x (cos⁡x+cos⁡3x)2cos⁡6x (cos⁡x+cos⁡3x).\frac{2 \sin 6x \, (\cos x + \cos 3x)}{2 \cos 6x \, (\cos x + \cos 3x)}. …

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