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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 2 · Set 2 · Q22
(a) Evaluate: (5 cos^2 60^°+4 sec^2 30^°- tan^2 45^°)/(sin^2 30^°+ cos^2 30^°)
(a)
Evaluate : \[\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}} \]
OR
(b)
Prove that : \[1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=\operatorname{cosec} \alpha \]
Marking-scheme solution
(a)
\(\displaystyle \frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
\(\displaystyle =\frac{5\left(\dfrac{1}{2}\right)^{2}+4\left(\dfrac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{1}\)
\(\displaystyle =\frac{67}{12}\)
OR
(b)
LHS \(\displaystyle =1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}\)
\(\displaystyle =1+\frac{(\operatorname{cosec} \alpha+1)(\operatorname{cosec} \alpha-1)}{(1+\operatorname{cosec} \alpha)}\)
\(\displaystyle =1+\operatorname{cosec} \alpha-1\)
\(\displaystyle =\operatorname{cosec} \alpha=\) RHS
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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.