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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 4 · Set 3 · Q21
Prove that: $\displaystyle \frac{\tan \theta}{1+\tan ^{2} \theta}+\frac{\cot \theta}{1+\cot ^{2} \theta}=2 \sin \theta \cos \theta$.Evaluate : $\displaystyle \frac{1-2 \tan ^{2} 30^{\circ}-\sec ^{2} 45^{\circ}}{\sin ^{2} 60^{\circ}}$
Prove that: $\displaystyle \frac{\tan \theta}{1+\tan ^{2} \theta}+\frac{\cot \theta}{1+\cot ^{2} \theta}=2 \sin \theta \cos \theta$.
Evaluate : $\displaystyle \frac{1-2 \tan ^{2} 30^{\circ}-\sec ^{2} 45^{\circ}}{\sin ^{2} 60^{\circ}}$
Marking-scheme solution
L.H.S. \(\displaystyle =\frac{\tan \theta}{\sec ^{2} \theta}+\frac{\cot \theta}{\operatorname{cosec}^{2} \theta}\)
\[=\frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta+\frac{\cos \theta}{\sin \theta} \times \sin ^{2} \theta
\]
\[=2 \sin \theta \cos \theta=\text { R.H.S. }
\]
\[\begin{aligned}
& =\frac{1-2 \times\left(\dfrac{1}{\sqrt{3}}\right)^{2}-(\sqrt{2})^{2}}{\left(\dfrac{\sqrt{3}}{2}\right)^{2}} \\
& =\frac{-20}{9}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.