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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 :

Solution

✅ Correct Option: 2
  1. Given: sin⁡θ+cos⁡θ=72\sin \theta + \cos \theta = \frac{\sqrt{7}}{2}
  1. 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)

  1. Now for sin⁡θ−cos⁡θ\sin \theta - \cos \theta, let's square:

(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

  1. From equation (1):

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

=1−34= 1 - \frac{3}{4}

=14= \frac{1}{4}

  1. Therefore:

sin⁡θ−cos⁡θ=±12\sin \theta - \cos \theta = \pm\frac{1}{2}

  1. 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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