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Question

If $\sin\theta - \cos\theta = \frac{\sqrt{3}}{2}$, then find the positive value of $\sin\theta + \cos\theta$.

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$\frac{\sqrt{5}}{2}$

Finding sin(theta) + cos(theta) Value

We are given the equation:
$\sin\theta - \cos\theta = \frac{\sqrt{3}}{2}$

We need to find the positive value of $\sin\theta + \cos\theta$. Let's denote this value as $x$.
$x = \sin\theta + \cos\theta$

Algebraic Manipulation

Square both the given equation and the equation for $x$:

  • Squaring the given equation: $(\sin\theta - \cos\theta)^2 = \left(\frac{\sqrt{3}}{2}\right)^2$ $\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = \frac{3}{4}$ Using the identity $\sin^2\theta + \cos^2\theta = 1$: $1 - 2\sin\theta\cos\theta = \frac{3}{4}$ $2\sin\theta\cos\theta = 1 - \frac{3}{4} = \frac{1}{4}$
  • Squaring the equation for $x$: $x^2 = (\sin\theta + \cos\theta)^2$ $x^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$ Using the identity $\sin^2\theta + \cos^2\theta = 1$: $x^2 = 1 + 2\sin\theta\cos\theta$

Calculating the Final Value

Substitute the value of $2\sin\theta\cos\theta$ found earlier into the equation for $x^2$:
$x^2 = 1 + \frac{1}{4}$
$x^2 = \frac{5}{4}$

Now, solve for $x$:
$x = \pm\sqrt{\frac{5}{4}}$
$x = \pm\frac{\sqrt{5}}{2}$

The question asks for the positive value.
Therefore, the positive value of $\sin\theta + \cos\theta$ is $\frac{\sqrt{5}}{2}$.

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