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NCERT Exemplar · Q76

Q.Match each item given under the column C1C_1 to its correct answer given under the column C2C_2. Column C1C_1:

(a) sin⁡(x+y)sin⁡(x−y)\sin(x + y)\sin(x - y);
(b) cos⁡(x+y)cos⁡(x−y)\cos(x + y)\cos(x - y);
(c) cot⁡(π4+θ)\cot\left(\dfrac{\pi}{4} + \theta\right);
(d) tan⁡(π4+θ)\tan\left(\dfrac{\pi}{4} + \theta\right). Column C2C_2:
(i) cos⁡2x−sin⁡2y\cos^2 x - \sin^2 y;
(ii) 1−tan⁡θ1+tan⁡θ\dfrac{1 - \tan\theta}{1 + \tan\theta};
(iii) 1+tan⁡θ1−tan⁡θ\dfrac{1 + \tan\theta}{1 - \tan\theta};
(iv) sin⁡2x−sin⁡2y\sin^2 x - \sin^2 y.
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Each expression in C1C_1 simplifies using product-to-sum formulas or compound-angle identities to match exactly one form in C2C_2: (a)↔(iv), (b)↔(i), (c)↔(ii), (d)↔(iii).

The heart of this matching exercise lies in recognizing two families of identities: the product-to-sum formulas for sine and cosine products, and the tangent addition formula for expressions involving π4\frac{\pi}{4}. The first two items transform products into differences of squares, while the last two exploit the fact that tan⁡π4=1\tan\frac{\pi}{4} = 1 to produce elegant rational forms.


Matching (a): sin⁡(x+y)sin⁡(x−y)\sin(x + y)\sin(x - y)

We apply the product-to-sum identity. Recall that

sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)].\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)].

  1. Set A=x+yA = x + y and B=x−yB = x - y. Then:

A−B=(x+y)−(x−y)=2y,A+B=(x+y)+(x−y)=2x.A - B = (x + y) - (x - y) = 2y, \quad A + B = (x + y) + (x - y) = 2x.

  1. Substitute:

sin⁡(x+y)sin⁡(x−y)=12[cos⁡2y−cos⁡2x].\sin(x + y)\sin(x - y) = \frac{1}{2}[\cos 2y - \cos 2x].

  1. Use the double-angle formulas cos⁡2α=1−2sin⁡2α\cos 2\alpha = 1 - 2\sin^2\alpha (or equivalently cos⁡2α=2cos⁡2α−1\cos 2\alpha = 2\cos^2\alpha - 1):

cos⁡2y=1−2sin⁡2y,cos⁡2x=1−2sin⁡2x.\cos 2y = 1 - 2\sin^2 y, \quad \cos 2x = 1 - 2\sin^2 x.

  1. Therefore:

12[(1−2sin⁡2y)−(1−2sin⁡2x)]=12[2sin⁡2x−2sin⁡2y]=sin⁡2x−sin⁡2y.\frac{1}{2}[(1 - 2\sin^2 y) - (1 - 2\sin^2 x)] = \frac{1}{2}[2\sin^2 x - 2\sin^2 y] = \sin^2 x - \sin^2 y.

This matches (iv).


Matching (b): cos⁡(x+y)cos⁡(x−y)\cos(x + y)\cos(x - y)

The product-to-sum identity for cosines is

cos⁡Acos⁡B=12[cos⁡(A−B)+cos⁡(A+B)].\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)].

  1. Again, A=x+yA = x + y, B=x−yB = x - y, so A−B=2yA - B = 2y and A+B=2xA + B = 2x.

  2. Substitute:

cos⁡(x+y)cos⁡(x−y)=12[cos⁡2y+cos⁡2x].\cos(x + y)\cos(x - y) = \frac{1}{2}[\cos 2y + \cos 2x].

  1. Now use the double-angle formula cos⁡2α=2cos⁡2α−1\cos 2\alpha = 2\cos^2\alpha - 1:

cos⁡2x=2cos⁡2x−1,cos⁡2y=2cos⁡2y−1.\cos 2x = 2\cos^2 x - 1, \quad \cos 2y = 2\cos^2 y - 1.

  1. Therefore:

12[(2cos⁡2x−1)+(2cos⁡2y−1)]=12[2cos⁡2x+2cos⁡2y−2]=cos⁡2x+cos⁡2y−1.\frac{1}{2}[(2\cos^2 x - 1) + (2\cos^2 y - 1)] = \frac{1}{2}[2\cos^2 x + 2\cos^2 y - 2] = \cos^2 x + \cos^2 y - 1.

  1. Rewrite using sin⁡2y=1−cos⁡2y\sin^2 y = 1 - \cos^2 y:

cos⁡2x+cos⁡2y−1=cos⁡2x−(1−cos⁡2y)=cos⁡2x−sin⁡2y.\cos^2 x + \cos^2 y - 1 = \cos^2 x - (1 - \cos^2 y) = \cos^2 x - \sin^2 y.

This matches (i).

Watch out

A common mistake is to stop at cos⁡2x+cos⁡2y−1\cos^2 x + \cos^2 y - 1 without recognizing it equals cos⁡2x−sin⁡2y\cos^2 x - \sin^2 y. Always check if the answer can be rewritten in the given forms.


Matching (c): cot⁡(π4+θ)\cot\left(\frac{\pi}{4} + \theta\right)

Cotangent is the reciprocal of tangent, so we first find tan⁡(π4+θ)\tan\left(\frac{\pi}{4} + \theta\right) and then invert.

  1. The tangent addition formula gives: …

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