CBSE 2025 · Region 5 · Set 1 · Q30 · 3 marks
Prove the following trigonometric identity : \[\frac{1+\operatorname{cosec} \mathrm{A}}{\operatorname{cosec} \mathrm{A}}=\frac{\cos ^{2} \mathrm{A}}{1-\sin \mathrm{A}} \]Let $\displaystyle 2 \mathrm{~A}+\mathrm{B}$ and $\displaystyle \mathrm{A}+2 \mathrm{~B}$ be acute angles such that $\displaystyle \sin (2 \mathrm{~A}+\mathrm{B})=\frac{\sqrt{3}}{2}$ and $\displaystyle \tan (\mathrm{A}+2 \mathrm{~B})=1$. Find the value of $\displaystyle \cot (4 \mathrm{~A}-7 \mathrm{~B})$.
Prove the following trigonometric identity : \[\frac{1+\operatorname{cosec} \mathrm{A}}{\operatorname{cosec} \mathrm{A}}=\frac{\cos ^{2} \mathrm{A}}{1-\sin \mathrm{A}} \]
Let $\displaystyle 2 \mathrm{~A}+\mathrm{B}$ and $\displaystyle \mathrm{A}+2 \mathrm{~B}$ be acute angles such that $\displaystyle \sin (2 \mathrm{~A}+\mathrm{B})=\frac{\sqrt{3}}{2}$ and $\displaystyle \tan (\mathrm{A}+2 \mathrm{~B})=1$. Find the value of $\displaystyle \cot (4 \mathrm{~A}-7 \mathrm{~B})$.
Marking-scheme solution
\[\begin{aligned}
\mathrm{LHS} & =\frac{1+\dfrac{1}{\sin \mathrm{~A}}}{\dfrac{1}{\sin \mathrm{~A}}} \\
& =\sin \mathrm{A}+1 \\
& =\frac{(\sin \mathrm{A}+1)(1-\sin \mathrm{A})}{1-\sin \mathrm{A}} \\
& =\frac{1-\sin ^{2} \mathrm{~A}}{1-\sin \mathrm{A}} \\
& =\frac{\cos ^{2} \mathrm{~A}}{1-\sin \mathrm{A}}=\mathrm{RHS}
\end{aligned}
\]
\[\begin{aligned}
& \sin (2 A+B)=\frac{\sqrt{3}}{2} \Rightarrow 2 A+B=60^{\circ} \quad \text { --- (1) } \\
& \tan (A+2 B)=1 \Rightarrow A+2 B=45^{\circ} \quad \text { --- (2) } \\
& \text { Solving (1) \& (2), we get } A=25^{\circ} \text { and } B=10^{\circ} \\
& \begin{aligned}
\cot (4 A-7 B) & =\cot 30^{\circ} \\
& =\sqrt{3}
\end{aligned}
\end{aligned}
\]
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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.