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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 4 · Set 3 · Q22
(A) Evaluate 2 sec^2 θ+3 cosec^2 θ-2 sin θ cos θ if θ=45^°.
(A)
Evaluate $\displaystyle 2 \sec ^{2} \theta+3 \operatorname{cosec}^{2} \theta-2 \sin \theta \cos \theta$ if $\displaystyle \theta=45^{\circ}$.
OR
(B)
If $\displaystyle \sin \theta-\cos \theta=0$, then find the value of $\displaystyle \sin ^{4} \theta+\cos ^{4} \theta$.
Marking-scheme solution
(A)
\[\begin{aligned}
& 2 \sec ^{2} 45^{\circ}+3 \operatorname{cosec}^{2} 45^{\circ}-2 \sin 45^{\circ} \cos 45^{\circ} \\
& =2(\sqrt{2})^{2}+3(\sqrt{2})^{2}-2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) \\
& =4+6-1=9
\end{aligned}
\]
(B)
\[\begin{aligned}
& \sin \theta-\cos \theta=0 \Rightarrow \sin \theta=\cos \theta \Rightarrow \tan \theta=1 \\
& \Rightarrow \theta=45^{\circ} \\
& \sin ^{4} 45^{\circ}+\cos ^{4} 45^{\circ}=\left(\frac{1}{\sqrt{2}}\right)^{4}+\left(\frac{1}{\sqrt{2}}\right)^{4} \\
& =\frac{1}{4}+\frac{1}{4}=\frac{1}{2}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.