Question : If $\sin (x+y) = \cos (x–y)$, then the value of $\cos^2 x$ is:
Option 1: $\frac{1}{2}$
Option 2: $3$
Option 3: $5$
Option 4: $\frac{1}{4}$
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Correct Answer: $\frac{1}{2}$
Solution : Given: $\sin (x+y) = \cos (x–y)$ If $\theta_1+\theta_2=90^{\circ}$, then If $\sin \theta _1=\cos \theta_2$. $\sin (x+y) = \cos (x–y)$ ⇒ $x+y+x–y=90^{\circ}$ ⇒ $2x=90^{\circ}$ ⇒ $x=45^{\circ}$ The value of $\cos^2 x=\cos^245^{\circ}$ = $(\frac{1}{\sqrt2})^2$ = $\frac{1}{2}$ Hence, the correct answer is $\frac{1}{2}$.
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Question : If $x\sin^{3}\theta +y\cos^{3}\theta=\sin\theta\cos\theta$ and $x\sin\theta-y\cos\theta=0$, then the value of $\left ( x^{2}+y^{2} \right )$ equals:
Option 1: $1$
Option 2: $\frac{1}{2}$
Option 3: $\frac{3}2$
Option 4: $2$
Question : If $\tan x = \frac{7}{5}$, the value of $\frac{9 \sin x – \frac{42}{5} \cos x}{15 \sin x + 21 \cos x}$ is:
Option 1: 0
Option 2: 1
Option 3: 0.1
Option 4: 0.5
Question : If $x=a(\sin\theta+\cos\theta), y=b(\sin\theta-\cos\theta)$, then the value of $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is:
Option 3: 2
Option 4: –2
Question : If $2\cot x=5$, then what is $\frac{2 \cos x-\sin x}{2 \cos x+\sin x}$ equal to?
Option 1: $\frac{3}{4}$
Option 2: $\frac{1}{3}$
Option 3: $\frac{5}{6}$
Option 4: $\frac{2}{3}$
Question : If $\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{2}{5}$, then the value of $\frac{1+\cot ^2 x}{1-\cot ^2 x}$ is:
Option 1: 2.25
Option 2: 1.45
Option 3: 3.75
Option 4: 5.25
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