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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 2 · Set 1 · Q24
If $\displaystyle \tan \theta+\frac{1}{\tan \theta}=2$, find the value of $\displaystyle \tan ^{2} \theta+\frac{1}{\tan ^{2} \theta}$.Prove that : $\displaystyle \sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$
If $\displaystyle \tan \theta+\frac{1}{\tan \theta}=2$, find the value of $\displaystyle \tan ^{2} \theta+\frac{1}{\tan ^{2} \theta}$.
Prove that : $\displaystyle \sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$
Marking-scheme solution
\(\displaystyle \left(\tan \theta+\frac{1}{\tan \theta}\right)^{2}=(2)^{2}\)
\(\displaystyle \Rightarrow \tan ^{2} \theta+\frac{1}{\tan ^{2} \theta}+2=4\)
\(\displaystyle \Rightarrow \tan ^{2} \theta+\frac{1}{\tan ^{2} \theta}=2\)
LHS \(\displaystyle =\sqrt{\frac{(1-\sin \theta)}{(1+\sin \theta)} \times \frac{(1-\sin \theta)}{(1-\sin \theta)}}\)
\(\displaystyle =\sqrt{\frac{(1-\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}\)
\(\displaystyle =\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}}\)
\(\displaystyle =\frac{(1-\sin \theta)}{\cos \theta}=\sec \theta-\tan \theta=\mathrm{RHS}\)
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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.