JIPMAT 2024 (QA) - If + =sqrt(7)2, then ( - ) is equal to : | PYQs + Solutions | AfterBoards
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JIPMAT 2024

Geometry
>
Trigonometry

Hard

If sin⁡θ+cos⁡θ=72\sin \theta+\cos \theta=\frac{\sqrt{7}}{2}, then (sin⁡θ−cos⁡θ)(\sin \theta-\cos \theta) is equal to :

Correct Option: 2
1) Given: sin⁡θ+cos⁡θ=72\sin \theta + \cos \theta = \frac{\sqrt{7}}{2}
2) Let's square this equation:
(sin⁡θ+cos⁡θ)2=74(\sin \theta + \cos \theta)^2 = \frac{7}{4}
sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=74\sin^2 \theta + 2\sin \theta\cos \theta + \cos^2 \theta = \frac{7}{4}
Since sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1:
1+2sin⁡θcos⁡θ=741 + 2\sin \theta\cos \theta = \frac{7}{4}
2sin⁡θcos⁡θ=342\sin \theta\cos \theta = \frac{3}{4} ...(1)
3) Now for sin⁡θ−cos⁡θ\sin \theta - \cos \theta, let's square: \newline (sin⁡θ−cos⁡θ)2(\sin \theta - \cos \theta)^2
=sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ= \sin^2 \theta - 2\sin \theta\cos \theta + \cos^2 \theta
=1−2sin⁡θcos⁡θ= 1 - 2\sin \theta\cos \theta
4) From equation (1): \newline (sin⁡θ−cos⁡θ)2(\sin \theta - \cos \theta)^2
=1−34= 1 - \frac{3}{4}
=14= \frac{1}{4}
5) Therefore: \newline sin⁡θ−cos⁡θ=±12\sin \theta - \cos \theta = \pm\frac{1}{2}
6) Since sin⁡θ+cos⁡θ=72\sin \theta + \cos \theta = \frac{\sqrt{7}}{2} is positive,
sin⁡θ−cos⁡θ=12\sin \theta - \cos \theta = \frac{1}{2}

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Given below are two statement:
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