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NCERT Exemplar · Q13

Q.If sin⁡(x+y)sin⁡(x−y)=a+ba−b\dfrac{\sin(x + y)}{\sin(x - y)} = \dfrac{a + b}{a - b}, then show that tan⁡xtan⁡y=ab\dfrac{\tan x}{\tan y} = \dfrac{a}{b}.

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Apply componendo-dividendo to the given ratio, then use the sum and difference formulas for sine to express everything in terms of tan⁡x\tan x and tan⁡y\tan y. The result tan⁡xtan⁡y=ab\frac{\tan x}{\tan y} = \frac{a}{b} follows directly.

The heart of this problem lies in recognizing that a ratio equation invites componendo-dividendo, and that sine sum/difference formulas naturally lead to tangent ratios. When you see sin⁡(x+y)\sin(x + y) and sin⁡(x−y)\sin(x - y) together, think: these can be expanded and their ratio will simplify beautifully when we add and subtract the equations.

The strategy is to manipulate the given ratio using componendo-dividendo (adding and subtracting numerators and denominators), then expand the sine terms using standard formulas. The cross-terms will cancel in just the right way to leave us with tangents.

Step-by-step derivation

  1. Apply componendo-dividendo to the given equation.

    We have sin⁡(x+y)sin⁡(x−y)=a+ba−b\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b}.

    By componendo-dividendo (if pq=rs\frac{p}{q} = \frac{r}{s}, then p+qp−q=r+sr−s\frac{p + q}{p - q} = \frac{r + s}{r - s}):

sin⁡(x+y)+sin⁡(x−y)sin⁡(x+y)−sin⁡(x−y)=(a+b)+(a−b)(a+b)−(a−b)=2a2b=ab\frac{\sin(x + y) + \sin(x - y)}{\sin(x + y) - \sin(x - y)} = \frac{(a + b) + (a - b)}{(a + b) - (a - b)} = \frac{2a}{2b} = \frac{a}{b}

  1. Expand the sine terms using sum and difference formulas.

    Recall that:

    • sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin(x + y) = \sin x \cos y + \cos x \sin y
    • sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin(x - y) = \sin x \cos y - \cos x \sin y

    Now compute the numerator:

sin⁡(x+y)+sin⁡(x−y)=(sin⁡xcos⁡y+cos⁡xsin⁡y)+(sin⁡xcos⁡y−cos⁡xsin⁡y)=2sin⁡xcos⁡y\sin(x + y) + \sin(x - y) = (\sin x \cos y + \cos x \sin y) + (\sin x \cos y - \cos x \sin y) = 2\sin x \cos y

And the denominator:

sin⁡(x+y)−sin⁡(x−y)=(sin⁡xcos⁡y+cos⁡xsin⁡y)−(sin⁡xcos⁡y−cos⁡xsin⁡y)=2cos⁡xsin⁡y\sin(x + y) - \sin(x - y) = (\sin x \cos y + \cos x \sin y) - (\sin x \cos y - \cos x \sin y) = 2\cos x \sin y

  1. Substitute back into our ratio.

    2sin⁡xcos⁡y2cos⁡xsin⁡y=ab\frac{2\sin x \cos y}{2\cos x \sin y} = \frac{a}{b} …

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