Question : If $4 \cos θ + 3 \sin θ = x$ and $4 \sin θ − 3 \cos θ = y$, find the value of $x^2 + y^2$.
Option 1: 16
Option 2: 9
Option 3: 25
Option 4: 1
Correct Answer: 25
Solution : Given: $(4 \cos θ + 3 \sin θ)^2 = x^2$ $(4 \sin θ - 3 \cos θ)^2 = y^2$ Now, $x^2 + y^2 = (4 \cos θ + 3 \sin θ)^2 + (4 \sin θ - 3 \cos θ)^2$ $= 16 \cos^2 θ + 24 \cos θ \sin θ + 9 \sin^2 θ + 16 \sin^2 θ - 24 \cos θ \sin θ + 9 \cos^2 θ$ $= 16(\cos^2 θ + \sin^2 θ) + 9(\cos^2 θ + \sin^2 θ)$ $ = (16\times1) + (9\times1$) [Since $\cos^2 θ + \sin^2 θ = 1$] $= 25$ Hence, the correct answer is 25.
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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: $\frac{(1+x)}{x}$
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 $\sin \theta+\cos \theta=\sqrt{2} \cos \theta$, 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 $\cos\theta=\frac{3}{5}$, then the value of $\sin\theta.\sec\theta.\tan\theta$ is:
Option 1: $\frac{9}{16}$
Option 2: $\frac{16}{9}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{4}{3}$
Question : The value of $\frac{2 \cos ^3 \theta-\cos \theta}{\sin \theta-2 \sin ^3 \theta}$ is:
Option 1: $\sec \theta$
Option 2: $\sin \theta$
Option 3: $\cot \theta$
Option 4: $\tan \theta$
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