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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 5 · Set 1 · Q25
If $\displaystyle \mathrm{a} \cos \theta+\mathrm{b} \sin \theta=\mathrm{m}$ and $\displaystyle \mathrm{a} \sin \theta-\mathrm{b} \cos \theta=\mathrm{n}$, then prove that $\displaystyle \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{m}^{2}+\mathrm{n}^{2}$.Prove that : \[\sqrt{\frac{\sec \mathrm{A}-1}{\sec \mathrm{~A}+1}}+\sqrt{\frac{\sec \mathrm{A}+1}{\sec \mathrm{~A}-1}}=2 \operatorname{cosec} \mathrm{~A} \]
If $\displaystyle \mathrm{a} \cos \theta+\mathrm{b} \sin \theta=\mathrm{m}$ and $\displaystyle \mathrm{a} \sin \theta-\mathrm{b} \cos \theta=\mathrm{n}$, then prove that $\displaystyle \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{m}^{2}+\mathrm{n}^{2}$.
Prove that : \[\sqrt{\frac{\sec \mathrm{A}-1}{\sec \mathrm{~A}+1}}+\sqrt{\frac{\sec \mathrm{A}+1}{\sec \mathrm{~A}-1}}=2 \operatorname{cosec} \mathrm{~A} \]
Marking-scheme solution
\[\begin{aligned}
& m^{2}+n^{2}=(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2} \\
& =a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\
& =a^{2}+b^{2}
\end{aligned}
\]
\[\begin{aligned}
& \text { LHS }=\frac{\sqrt{\sec \mathrm{A}-1}}{\sqrt{\sec \mathrm{~A}+1}}+\frac{\sqrt{\sec \mathrm{A}+1}}{\sqrt{\sec \mathrm{~A}-1}} \\
& =\frac{\sec \mathrm{A}-1+\sec \mathrm{A}+1}{\sqrt{\sec ^{2} A-1}}
\end{aligned}
\]
\[\begin{aligned}
& =\frac{2 \sec A}{\tan A} \\
& =2 \operatorname{cosec} A=\text { RHS }
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.