Question

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

• A. \(\frac{5}{\sqrt{34}}\)
• B. \(-\frac{5}{\sqrt{34}}\)
• C. \(-\frac{\sqrt{34}}{25}\)
• D. \(\frac{\sqrt{34}}{5}\)

Hint:

Use \((\sin\theta+\cos\theta)^2\) and \((\sin\theta-\cos\theta)^2\). Their difference depends on \(2\sin\theta\cos\theta\).

Answer 1

The Correct Option is D

We are given \[ \sin\theta-\cos\theta=\frac{4}{5}. \] We need to find \[ \sin\theta+\cos\theta. \] Let \[ S=\sin\theta+\cos\theta. \] Now square both: \[ (\sin\theta+\cos\theta)^2 = \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta. \] Using \[ \sin^2\theta+\cos^2\theta=1, \] we get \[ S^2=1+2\sin\theta\cos\theta. \] Now square the given expression: \[ (\sin\theta-\cos\theta)^2=\left(\frac{4}{5}\right)^2. \] \[ \sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta=\frac{16}{25}. \] \[ 1-2\sin\theta\cos\theta=\frac{16}{25}. \] So, \[ 2\sin\theta\cos\theta=1-\frac{16}{25}. \] \[ 2\sin\theta\cos\theta=\frac{9}{25}. \] Now substitute in \(S^2\): \[ S^2=1+\frac{9}{25}. \] \[ S^2=\frac{25}{25}+\frac{9}{25}. \] \[ S^2=\frac{34}{25}. \] Taking square root: \[ S=\frac{\sqrt{34}}{5}. \] Therefore, \[ \sin\theta+\cos\theta=\frac{\sqrt{34}}{5}. \]