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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 2 · Set 1 · Q22
If $\displaystyle 4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}+\sin ^{2} 60^{\circ}+\mathrm{p}=\frac{3}{4}$, then find the value of p.If $\displaystyle \cos \mathrm{A}+\cos ^{2} \mathrm{~A}=1$, then find the value of $\displaystyle \sin ^{2} \mathrm{~A}+\sin ^{4} \mathrm{~A}$.
If $\displaystyle 4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}+\sin ^{2} 60^{\circ}+\mathrm{p}=\frac{3}{4}$, then find the value of p.
If $\displaystyle \cos \mathrm{A}+\cos ^{2} \mathrm{~A}=1$, then find the value of $\displaystyle \sin ^{2} \mathrm{~A}+\sin ^{4} \mathrm{~A}$.
Marking-scheme solution
\[\begin{aligned}
& 4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}+\sin ^{2} 60^{\circ}+\mathrm{p}=\frac{3}{4} \\
& \Rightarrow 4(1)^{2}-(2)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}+\mathrm{p}=\frac{3}{4} \\
& \Rightarrow 4-4+\frac{3}{4}+\mathrm{p}=\frac{3}{4} \\
& \Rightarrow \mathrm{p}=0
\end{aligned}
\]
\[\begin{aligned}
\cos A+\cos ^{2} A=1 & \Rightarrow \cos A=1-\cos ^{2} A=\sin ^{2} A \\
\therefore \sin ^{2} A+\sin ^{4} A & =\cos A+\cos ^{2} A\left(\because \sin ^{2} A=\cos A\right) \\
& =1
\end{aligned}
\]
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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.