Question : If $x=\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}$, then the value of $\frac{1-\cos \theta+\sin \theta}{(1+\sin \theta)}$ is:
Option 1: $\frac{x}{(1+x)}$
Option 2: $x$
Option 3: $\frac{1}{x}$
Option 4: $\frac{(1+x)}{x}$
Correct Answer: $x$
Solution : Given expression, $x=\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}$ Multiply and divide by $1-\cos \theta+\sin \theta$, ⇒ $x=\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}\times\frac{1-\cos \theta+\sin \theta}{1-\cos \theta+\sin \theta}$ ⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\cos \theta+\sin \theta)({1-\cos \theta+\sin \theta})}$ ⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)^2-(\cos^2\theta)}$ ⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)^2-(1-\sin^2\theta)}$ [As $\cos^2\theta+\sin^2\theta=1$] ⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)^2-(1-\sin\theta)(1+\sin\theta)}$ ⇒ $=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)[(1+\sin\theta)-(1-\sin\theta)]}$ ⇒ $x=\frac{2\sin\theta(1-\cos \theta+\sin \theta)}{(1+\sin\theta)(2\sin\theta)}$ ⇒ $\frac{(1-\cos \theta+\sin \theta)}{(1+\sin\theta)}=x$ Hence, the correct answer is $x$.
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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: $\frac{1}{2}\cos \theta$
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: $\frac{\sqrt{7}}{3}$
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: $\sqrt{2}$
Question : If $\cos \theta+\sec \theta=2$, then the value of $\sin ^6 \theta+\cos ^6 \theta$ is:
Option 1: $\frac{1}{3}$
Option 2: $0$
Option 4: $\frac{1}{2}$
Question : If $1 + \sin^2 θ - 3\sinθ \cosθ = 0$, then the value of $\cotθ$ is:
Option 1: $0$
Option 2: $2$
Option 3: $\frac{1}{2}$
Option 4: $\frac{1}{3}$
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