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JIPMAT 2025 PYQsQuantitative Ability. Free, no login required.

Geometry

past year questions, JIPMAT 2025

Trigonometry

past year questions, Geometry, JIPMAT 2025

Hard

If sin⁡θ+cos⁡θsin⁡θ−cos⁡θ=3\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = 3, then value of sin⁡4θ−cos⁡4θ\sin^4\theta - \cos^4\theta is

Solution

✅ Correct Option: 3

Given: sin⁡θ+cos⁡θsin⁡θ−cos⁡θ=3\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = 3

Cross multiply to eliminate the fraction:

sin⁡θ+cos⁡θ=3(sin⁡θ−cos⁡θ)\sin\theta + \cos\theta = 3(\sin\theta - \cos\theta)

Expand and collect like terms:

sin⁡θ+cos⁡θ=3sin⁡θ−3cos⁡θ\sin\theta + \cos\theta = 3\sin\theta - 3\cos\theta

sin⁡θ−3sin⁡θ=−cos⁡θ−3cos⁡θ\sin\theta - 3\sin\theta = -\cos\theta - 3\cos\theta

−2sin⁡θ=−4cos⁡θ-2\sin\theta = -4\cos\theta

sin⁡θ=2cos⁡θ\sin\theta = 2\cos\theta

Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1:

(2cos⁡θ)2+cos⁡2θ=1(2\cos\theta)^2 + \cos^2\theta = 1

4cos⁡2θ+cos⁡2θ=14\cos^2\theta + \cos^2\theta = 1

5cos⁡2θ=15\cos^2\theta = 1

cos⁡2θ=15\cos^2\theta = \frac{1}{5}

Therefore: sin⁡2θ=4cos⁡2θ=45\sin^2\theta = 4\cos^2\theta = \frac{4}{5}

Apply the difference of squares formula a4−b4=(a2+b2)(a2−b2)a^4 - b^4 = (a^2 + b^2)(a^2 - b^2):

sin⁡4θ−cos⁡4θ=(sin⁡2θ+cos⁡2θ)(sin⁡2θ−cos⁡2θ)\sin^4\theta - \cos^4\theta = (\sin^2\theta + \cos^2\theta)(\sin^2\theta - \cos^2\theta)

Substitute our values:

=(1)(sin⁡2θ−cos⁡2θ)= (1)(\sin^2\theta - \cos^2\theta)

=45−15= \frac{4}{5} - \frac{1}{5}

=35= \frac{3}{5}

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