CBSE 2025 · Region 2 · Set 1 · Q27 · 3 marks
Prove that : $\displaystyle \left(1+\frac{1}{\tan ^{2} \theta}\right)\left(1+\frac{1}{\cot ^{2} \theta}\right)=\frac{1}{\sin ^{2} \theta-\sin ^{4} \theta}$Prove that: $\displaystyle \sqrt{\frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1}}+\sqrt{\frac{\operatorname{cosec} \theta+1}{\operatorname{cosec} \theta-1}}=2 \sec \theta$
Prove that : $\displaystyle \left(1+\frac{1}{\tan ^{2} \theta}\right)\left(1+\frac{1}{\cot ^{2} \theta}\right)=\frac{1}{\sin ^{2} \theta-\sin ^{4} \theta}$
Prove that: $\displaystyle \sqrt{\frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1}}+\sqrt{\frac{\operatorname{cosec} \theta+1}{\operatorname{cosec} \theta-1}}=2 \sec \theta$
Marking-scheme solution
\[\begin{aligned}
\mathrm{LHS} & =\left(1+\cot ^{2} \theta\right)\left(1+\tan ^{2} \theta\right) \\
& =\operatorname{cosec}^{2} \theta \cdot \sec ^{2} \theta \\
& =\frac{1}{\sin ^{2} \theta} \cdot \frac{1}{\cos ^{2} \theta} \\
& =\frac{1}{\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)} \\
& =\frac{1}{\sin ^{2} \theta-\sin ^{4} \theta}=\mathrm{RHS}
\end{aligned}
\]
\[\begin{aligned}
\mathrm{LHS} & =\frac{\operatorname{cosec} \theta-1+\operatorname{cosec} \theta+1}{\sqrt{(\operatorname{cosec} \theta+1)(\operatorname{cosec} \theta-1)}} \\
& =\frac{2 \operatorname{cosec} \theta}{\sqrt{\left(\operatorname{cosec}^{2} \theta-1\right)}} \\
& =\frac{2 \operatorname{cosec} \theta}{\cot \theta} \\
& =2 \sec \theta=\mathrm{RHS}
\end{aligned}
\]
Introduction to TrigonometryTrigonometric IdentitiesApplyshort_answerhard
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.