CBSE 2025 · Region 2 · Set 3 · Q26 · 3 marks
Prove that : $\displaystyle \sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}=\tan \theta+\cot \theta$If $\displaystyle \operatorname{cosec} \theta=\mathrm{x}+\frac{1}{4 \mathrm{x}}$, prove that $\displaystyle \operatorname{cosec} \theta+\cot \theta=2 \mathrm{x}$ or $\displaystyle \frac{1}{2 \mathrm{x}}$.
Prove that : $\displaystyle \sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}=\tan \theta+\cot \theta$
If $\displaystyle \operatorname{cosec} \theta=\mathrm{x}+\frac{1}{4 \mathrm{x}}$, prove that $\displaystyle \operatorname{cosec} \theta+\cot \theta=2 \mathrm{x}$ or $\displaystyle \frac{1}{2 \mathrm{x}}$.
Marking-scheme solution
LHS \(\displaystyle =\sqrt{\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}}\)
\[\begin{aligned}
& =\frac{1}{\sin \theta \cdot \cos \theta} \\
& =\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta} \\
& =\frac{\sin ^{2} \theta}{\sin \theta \cdot \cos \theta}+\frac{\cos ^{2} \theta}{\sin \theta \cdot \cos \theta} \\
& =\tan \theta+\cot \theta=\mathrm{RHS}
\end{aligned}
\]
\[\begin{aligned}
& \begin{aligned}
\cot ^{2} \theta=\operatorname{cosec}^{2} \theta-1 & =\left(x+\frac{1}{4 x}\right)^{2}-1 \\
& =\left(x-\frac{1}{4 x}\right)^{2}
\end{aligned} \\
& \Rightarrow \cot \theta=\left(x-\frac{1}{4 x}\right) \text { or }\left(\frac{1}{4 x}-x\right) \\
& \operatorname{cosec} \theta+\cot \theta=\left(x+\frac{1}{4 x}\right)+\left(x-\frac{1}{4 x}\right) \text { or }\left(x+\frac{1}{4 x}\right)+\left(\frac{1}{4 x}-x\right) \\
& \quad=2 x \text { or } \frac{1}{2 x}
\end{aligned}
\]
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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.