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 $2\cot x=5$, then what is $\frac{2 \cos x-\sin x}{2 \cos x+\sin x}$ equal to?
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:
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