Question : If $\sec x+\cos x=\frac{5}{2}$, where $x$ lies between $0^{\circ}$ and $90^{\circ}$, then what is the value of $\sin ^2 x$ ?
Option 1: $\frac{3}{4}$
Option 2: $\frac{1}{2}$
Option 3: $1$
Option 4: $\frac{1}{4}$
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Correct Answer: $\frac{3}{4}$
Solution :
Given: $\sec x + \cos x = \frac{5}{2}$
Now, $\cos\theta = \frac{1}{\sec\theta}$
⇒ $\frac{1}{\cos x} + \cos x = \frac{5}{2}$
The above equation satisfies for $x = 60^\circ$
Thus, $\sin^2x = \sin^2 60^\circ= (\frac{\sqrt3}{2})^2$
$\therefore \sin^2x = \frac{3}{4}$
Hence, the correct answer is $\frac{3}{4}$.
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