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Mathematics · 2024 · 2 marks
CBSE 2024 · Region 2 · Set 1 · Q23
(a) Evaluate: 2 sin^2 30^° sec 60^°+ tan^2 60^°.
(a)
Evaluate : $\displaystyle 2 \sin ^{2} 30^{\circ} \sec 60^{\circ}+\tan ^{2} 60^{\circ}$.
OR
(b)
If $\displaystyle 2 \sin (\mathrm{~A}+\mathrm{B})=\sqrt{3}$ and $\displaystyle \cos (\mathrm{A}-\mathrm{B})=1$, then find the measures of angles A and $\displaystyle \mathrm{B} .0 \leq \mathrm{A}, \mathrm{B},(\mathrm{A}+\mathrm{B}) \leq 90^{\circ}$.
Marking-scheme solution
(a)
\[\begin{aligned}
& 2 \sin ^{2} 30^{\circ} \sec 60^{\circ}+\tan ^{2} 60^{\circ} \\
& =2 \times\left(\frac{1}{2}\right)^{2} \times 2+(\sqrt{3})^{2} \\
& =4
\end{aligned}
\]
OR
(b)
\(\displaystyle \sin (\mathrm{A}+\mathrm{B})=\frac{\sqrt{3}}{2} \Rightarrow \mathrm{~A}+\mathrm{B}=60^{\circ} \ldots(1)\)
\[\cos (\mathrm{A}-\mathrm{B})=1 \Rightarrow \mathrm{~A}-\mathrm{B}=0^{\circ} \ldots(2)
\]
Solving ($\displaystyle 1$) and ($\displaystyle 2$), we get \(\displaystyle \mathrm{A}=\mathrm{B}=30^{\circ}\)
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.