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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