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Worked Examples · Example 13

Q.Prove that sin⁡(x+y)sin⁡(x−y)=tan⁡x+tan⁡ytan⁡x−tan⁡y\dfrac{\sin(x+y)}{\sin(x-y)} = \dfrac{\tan x + \tan y}{\tan x - \tan y}.

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This identity is proved by expanding sin⁡(x+y)\sin(x+y) and sin⁡(x−y)\sin(x-y) using sum-to-product formulas, then dividing numerator and denominator by cos⁡xcos⁡y\cos x \cos y to rewrite everything in terms of tan⁡x\tan x and tan⁡y\tan y.

We start with the left-hand side and aim to transform it into the right-hand side. The key is to express the sines in terms of sines and cosines of xx and yy, then introduce tangents by dividing by cos⁡xcos⁡y\cos x \cos y.

  1. Expand the sines using the sum/difference formulas. Recall: sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A \cos B + \cos A \sin B sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A \cos B - \cos A \sin B So:

sin⁡(x+y)sin⁡(x−y)=sin⁡xcos⁡y+cos⁡xsin⁡ysin⁡xcos⁡y−cos⁡xsin⁡y\frac{\sin(x+y)}{\sin(x-y)} = \frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y}

  1. Divide numerator and denominator by cos⁡xcos⁡y\cos x \cos y. Why? Because sin⁡xcos⁡x=tan⁡x\frac{\sin x}{\cos x} = \tan x and sin⁡ycos⁡y=tan⁡y\frac{\sin y}{\cos y} = \tan y. This step converts the expression into tangents.

=sin⁡xcos⁡ycos⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡ysin⁡xcos⁡ycos⁡xcos⁡y−cos⁡xsin⁡ycos⁡xcos⁡y= \frac{ \frac{\sin x \cos y}{\cos x \cos y} + \frac{\cos x \sin y}{\cos x \cos y} }{ \frac{\sin x \cos y}{\cos x \cos y} - \frac{\cos x \sin y}{\cos x \cos y} }

  1. Simplify each term.
    • sin⁡xcos⁡ycos⁡xcos⁡y=sin⁡xcos⁡x=tan⁡x\frac{\sin x \cos y}{\cos x \cos y} = \frac{\sin x}{\cos x} = \tan x
    • cos⁡xsin⁡ycos⁡xcos⁡y=sin⁡ycos⁡y=tan⁡y\frac{\cos x \sin y}{\cos x \cos y} = \frac{\sin y}{\cos y} = \tan y So the expression becomes:

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