Question:medium

If \(\sin\theta-\cos\theta=\frac{4}{5}\), then the value of \(\sin\theta+\cos\theta\) is

Show Hint

Use \((a+b)^2+(a-b)^2=2(a^2+b^2)\) for quick calculation.
  • \(\frac{5}{\sqrt{34}}\)
  • \(-\frac{5}{\sqrt{34}}\)
  • \(-\frac{\sqrt{34}}{25}\)
  • \(\frac{\sqrt{34}}{5}\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
We are given the value of $\sin\theta - \cos\theta$ and asked to find the value of $\sin\theta + \cos\theta$. This can be solved by squaring both expressions and using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$.
Step 2: Key Formula or Approach:
We use the algebraic identity $(a+b)^2 + (a-b)^2 = 2(a^2+b^2)$. Applying this to trigonometry: \[ (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2(\sin^2\theta + \cos^2\theta) = 2(1) = 2 \] Step 3: Detailed Explanation:
Let the unknown value be $y = \sin\theta + \cos\theta$. We are given that $\sin\theta - \cos\theta = \frac{4}{5}$. Using the identity from Step 2: \[ (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2 \] Substitute the known values: \[ y^2 + \left(\frac{4}{5}\right)^2 = 2 \] \[ y^2 + \frac{16}{25} = 2 \] Now, solve for $y^2$: \[ y^2 = 2 - \frac{16}{25} \] \[ y^2 = \frac{50}{25} - \frac{16}{25} = \frac{34}{25} \] Take the square root of both sides: \[ y = \pm \sqrt{\frac{34}{25}} = \pm \frac{\sqrt{34}}{5} \] The options include the positive value $\frac{\sqrt{34}}{5}$.
Step 4: Final Answer:
The value of $\sin\theta + \cos\theta$ is $\pm\frac{\sqrt{34}}{5}$. Since $\frac{\sqrt{34}}{5}$ is one of the options, it is the correct answer. Therefore, option (D) is correct.
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