Question

\(24\left(\sin^2 60^\circ + \cos^2 30^\circ + \cot^2 60^\circ + \csc^2 45^\circ\right) =\)

• A. \(23\)
• B. \(92\)
• C. \(46\)
• D. \(69\)

Hint:

When multiplying a sum of fractions by a constant, it often helps to multiply each term separately first, then add. Choosing the multiplier (here 24) as the LCD of all denominators (4, 3, etc.) is deliberate in exam problems to produce a clean integer answer.

Answer 1

The Correct Option is B

Step 1: Note the standard values.
Recall $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\cot 60^\circ = \frac{1}{\sqrt{3}}$, and $\csc 45^\circ = \sqrt{2}$. We will square each.

Step 2: Square the first two terms.
\[ \sin^2 60^\circ = \frac{3}{4}, \qquad \cos^2 30^\circ = \frac{3}{4} \]

Step 3: Square the last two terms.
\[ \cot^2 60^\circ = \frac{1}{3}, \qquad \csc^2 45^\circ = 2 \]

Step 4: Add the four values.
Use a common denominator of $12$. \[ \frac{9}{12} + \frac{9}{12} + \frac{4}{12} + \frac{24}{12} = \frac{46}{12} = \frac{23}{6} \]

Step 5: Multiply by $24$.
\[ 24\times \frac{23}{6} = 4\times 23 = 92 \]

Step 6: State the result.
So the whole expression equals $92$. Therefore \[ \boxed{92} \]