Question : If $\operatorname{cos} \theta+\operatorname{sin} \theta=\sqrt{2} \operatorname{cos} \theta$, find the value of $(\cos \theta-\operatorname{sin} \theta)$
Option 1: $\sqrt{2} \sin \theta$
Option 2: $\sqrt{2} \cos \theta$
Option 3: $\frac{1}{\sqrt{2}} \sin \theta$
Option 4: $\frac{1}{2}\cos \theta$
Correct Answer: $\sqrt{2} \sin \theta$
Solution : Given: $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$ Squaring both sides, we get: ⇒ $\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta = 2\cos^2 \theta$ ⇒ $\cos^2 \theta - \sin^2 \theta = 2\cos \theta \sin \theta$ ⇒$(\cos \theta + \sin \theta)(\cos \theta - \sin \theta) = 2\cos \theta \sin \theta$ Putting $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$, we get: $\sqrt{2}\cos \theta (\cos \theta - \sin \theta) = 2\cos \theta \sin \theta$ ⇒ $\cos \theta - \sin \theta = \sqrt{2} \sin \theta$ Hence, the correct answer is $\sqrt{2} \sin \theta$.
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Question : If $\sin \theta+\cos \theta=\sqrt{2} \cos \theta$, then find $\frac{\sin \theta-\cos \theta}{\sin \theta}$:
Question : If $\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4,0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\sec \theta+\operatorname{cosec} \theta+\cot \theta) ?$
Question : If $\sin \theta+\cos \theta=\frac{\sqrt{11}}{3}$, then the value of $(\cos \theta-\sin \theta)$ is:
Question : If $\sin \theta+\cos \theta=\frac{\sqrt{11}}{3}$, then what is $\sin \theta-\cos \theta$?
Question : If $\cos \theta-\sin \theta=\sqrt{2} \sin \theta$, then $(\cos \theta+\sin \theta)$ is:
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