Question : If $\sin \theta+\cos \theta=\sqrt{2} \cos \theta$, then find $\frac{\sin \theta-\cos \theta}{\sin \theta}$:
Option 1: $-\sqrt{2}$
Option 2: $-1$
Option 3: $1$
Option 4: $\sqrt{2}$
Correct Answer: $-\sqrt{2}$
Solution :
Given,
$\sin \theta+\cos \theta=\sqrt{2} \cos \theta$
Squaring both sides, we get
$\sin^2θ+\cos^2θ+2\sinθ\cosθ=2\cos^2θ$
⇒ $\sin^2θ−\cos^2θ+2\sinθ\cosθ=0$
Subtracting $2\sin^2θ$ both sides, we have
⇒ $−\sin^2θ−\cos^2θ+2\sinθ\cosθ=−2\sin^2θ$
⇒ $\sin^2θ+\cos^2θ−2\sinθ\cosθ=2\sin^2θ$
⇒ $(\cosθ−\sinθ)^2=2\sin^2θ$
⇒ $\cosθ−\sinθ=\sqrt2\sinθ$
⇒ $\sinθ-\cosθ=-\sqrt2\sinθ$
⇒ $\frac{\sin \theta-\cos \theta}{\sin \theta}=\frac{-\sqrt2\sinθ}{\sinθ}$
$=-\sqrt2$
Hence, the correct answer is $-\sqrt2$
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