CBSE 2025 · Region 2 · Set 2 · Q26 · 3 marks
If $\displaystyle \tan \theta+\sin \theta=\mathrm{m}$ and $\displaystyle \tan \theta-\sin \theta=\mathrm{n}$, then prove that $\displaystyle \mathrm{m}^{2}-\mathrm{n}^{2}=4 \sqrt{\mathrm{mn}}$.Prove that : $\displaystyle \frac{\cot \mathrm{A}-1}{2-\sec ^{2} \mathrm{~A}}=\frac{\cot \mathrm{A}}{1+\tan \mathrm{A}}$
If $\displaystyle \tan \theta+\sin \theta=\mathrm{m}$ and $\displaystyle \tan \theta-\sin \theta=\mathrm{n}$, then prove that $\displaystyle \mathrm{m}^{2}-\mathrm{n}^{2}=4 \sqrt{\mathrm{mn}}$.
Prove that : $\displaystyle \frac{\cot \mathrm{A}-1}{2-\sec ^{2} \mathrm{~A}}=\frac{\cot \mathrm{A}}{1+\tan \mathrm{A}}$
Marking-scheme solution
\[\begin{aligned}
\mathrm{LHS} & =\mathrm{m}^{2}-\mathrm{n}^{2} \\
& =(\tan \theta+\sin \theta)^{2}-(\tan \theta-\sin \theta)^{2} \\
& =4 \tan \theta \sin \theta \\
& =4 \sqrt{\tan ^{2} \theta \cdot \sin ^{2} \theta} \\
& =4 \sqrt{\tan ^{2} \theta\left(1-\cos ^{2} \theta\right)} \\
& =4 \sqrt{\tan ^{2} \theta-\sin ^{2} \theta} \\
& =4 \sqrt{(\tan \theta+\sin \theta)(\tan \theta-\sin \theta)} \\
& =4 \sqrt{\mathrm{mn}}=\mathrm{RHS}
\end{aligned}
\]
\[\begin{aligned}
\mathrm{LHS} & =\frac{\dfrac{1}{\tan \mathrm{~A}}-1}{2-\left(1+\tan ^{2} A\right)} \\
& =\frac{\dfrac{1-\tan \mathrm{A}}{\tan \mathrm{~A}}}{2-1-\tan ^{2} A} \\
& =\frac{1-\tan \mathrm{A}}{\tan \mathrm{~A}\left(1-\tan ^{2} \mathrm{~A}\right)} \\
& =\frac{1}{\tan \mathrm{~A}(1+\tan \mathrm{A})}
\end{aligned}
\]
\(\displaystyle =\frac{\cot \mathrm{A}}{1+\tan \mathrm{A}}=\mathrm{RHS}\)
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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.