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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 5 · Set 2 · Q22
(a) For acute angles A and B, if sec (2 A - B )=√ 2 and cosec ( A + B )=2, then find the values of A and B.
(a)
For acute angles A and B, if sec $\displaystyle (2 \mathrm{~A}-\mathrm{B})=\sqrt{2}$ and $\displaystyle \operatorname{cosec}(\mathrm{A}+\mathrm{B})=2$, then find the values of A and B.
OR
(b)
Evaluate : $\displaystyle \frac{2 \cos 30^{\circ}-\cot ^{3} 60^{\circ}}{\tan 30^{\circ}}$
Marking-scheme solution
$\displaystyle \sec (2 A-B)=\sqrt{2} \Rightarrow 2 A-B=45^{\circ}$
$\displaystyle \operatorname{cosec}(\mathrm{A}+\mathrm{B})=2 \Rightarrow \mathrm{~A}+\mathrm{B}=30^{\circ}$
On solving, $\displaystyle \mathrm{A}=25^{\circ}, \mathrm{B}=5^{\circ}$
Evaluate: $\displaystyle \frac{2 \cos 30^{\circ}-\cot ^{3} 60^{\circ}}{\tan 30^{\circ}}$
\[\begin{array}{l}
\frac{2 \cos 30^{\circ}-\cot ^{3} 60^{\circ}}{\tan 30^{\circ}}=\frac{2 \times \dfrac{\sqrt{3}}{2}-\left(\dfrac{1}{\sqrt{3}}\right)^{3}}{\dfrac{1}{\sqrt{3}}} \\
=\frac{8}{3}
\end{array}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.