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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 5 · Set 2 · Q23
(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 $\displaystyle \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{m}^{2}+\mathrm{n}^{2}$.(b)Prove that : \[\sqrt{\frac{\sec \mathrm{A}-1}{\sec \mathrm{~A}+1}}+\sqrt{\frac{\sec \mathrm{A}+1}{\sec \mathrm{~A}-1}}=2 \operatorname{cosec} \mathrm{~A} \]
(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 $\displaystyle \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{m}^{2}+\mathrm{n}^{2}$.
(b)
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
\[\begin{aligned}
& m^{2}+n^{2}=(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2} \\
& =a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\
& =a^{2}+b^{2}
\end{aligned}
\]
LHS \(\displaystyle =\frac{\sqrt{\boldsymbol{\operatorname { s e c }} \mathbf{A}-\mathbf{1}}}{\sqrt{\boldsymbol{\operatorname { s e c }} \mathbf{~}} \mathbf{1}+\mathbf{1}}+\frac{\sqrt{\boldsymbol{s e c} \mathbf{A}+\mathbf{1}}}{\sqrt{\boldsymbol{s e c} \mathbf{A}-\mathbf{1}}}\)
\[\begin{aligned}
& =\frac{\sec A-1+\sec A+1}{\sqrt{\sec ^{2} A-1}} \\
& =\frac{2 \sec A}{\tan A} \\
& =2 \operatorname{cosec} A=R H S
\end{aligned}
\]
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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.