Question : If $\sin \theta+\cos \theta=\frac{\sqrt{11}}{3}$, then the value of $(\cos \theta-\sin \theta)$ is:
Option 1: $\frac{\sqrt{5}}{3}$
Option 2: $\frac{7}{3}$
Option 3: $\frac{5}{3}$
Option 4: $\frac{\sqrt{7}}{3}$
Correct Answer: $\frac{\sqrt{7}}{3}$
Solution :
$\sin \theta+\cos \theta=\frac{\sqrt{11}}{3}$
Squaring,
$(\sin \theta+\cos \theta)^2=(\frac{\sqrt{11}}{3})^2$
⇒ $\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = \frac{11}{9}$
⇒ $1+2\sin \theta \cos \theta = \frac{11}{9}$
⇒ $2\sin \theta \cos \theta = \frac{11-9}{9}$
⇒ $2\sin \theta \cos \theta = \frac{2}{9}$ -----------(1)
$(\cos \theta-\sin \theta)^2$
$= \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta $
$= 1-2\sin \theta \cos \theta$
$=1-\frac{2}{9}$
$=\frac{7}{9}$
$\cos \theta-\sin \theta = \sqrt{\frac{7}{9}} = \frac{\sqrt7}{3}$
Hence, the correct answer is $ \frac{\sqrt7}{3}$.
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