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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 5 · Set 1 · Q24 (a)
Evaluate: (sin^3 60^°- tan 30^°)/(cos^2 45^°)
Evaluate: $\displaystyle \frac{\sin ^{3} 60^{\circ}-\tan 30^{\circ}}{\cos ^{2} 45^{\circ}}$
Marking-scheme solution
\[\begin{aligned}
& \frac{\sin ^{3} 60^{\circ}-\tan 30^{\circ}}{\cos ^{2} 45^{\circ}}=\frac{\left(\dfrac{\sqrt{3}}{2}\right)^{3}-\dfrac{1}{\sqrt{3}}}{\left(\dfrac{1}{\sqrt{2}}\right)^{2}} \\
& =\frac{1}{4 \sqrt{3}} \text { or } \frac{\sqrt{3}}{12}
\end{aligned}
\]
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