Question : If $\sin (x - y) = \frac{1}2$ and $\cos (x + y) = \frac{1}2$, then what is the value of $\sin x \cos x + 2\sin^2x + cos^3x \sec x$?
Option 1: $2$
Option 2: $\sqrt{2}+1$
Option 3: $1$
Option 4: $\frac{3}{4}$
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Correct Answer: $2$
Solution : $\sin (x - y) = \frac{1}2=\sin30^\circ$ $⇒(x-y)=30^\circ$---(1) $\cos (x + y) = \frac{1}2=\cos60^\circ$ $⇒(x+y)=60^\circ$---(2) Solving equation 1 and 2, we get, ⇒ $x=45^\circ$ Now, Putting the value of $x$, we get: $\sin x\cos x + 2\sin^2x + \cos^3 x \sec x$ $=\sin 45^\circ\cos45^\circ + 2\sin^2 45^\circ + \cos^3 45^\circ \sec 45^\circ$ $=\frac{1}{\sqrt2}\times \frac{1}{\sqrt2}+2\times\frac{1}{2}+\frac{1}{2\sqrt2}\times\sqrt2$ $=\frac{1}{2}+1+\frac{1}{2}$ $= 1 + 1$ $= 2$ Hence, the correct answer is $2$.
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