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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 1 · Set 3 · Q21
If $\displaystyle \sin \theta+\sin ^{2} \theta=1$, then prove that $\displaystyle \cos ^{2} \theta+\cos ^{4} \theta=1$.If $\displaystyle \tan \theta=\frac{1}{\sqrt{7}}$, then show that $\displaystyle \frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{3}{4}$.
If $\displaystyle \sin \theta+\sin ^{2} \theta=1$, then prove that $\displaystyle \cos ^{2} \theta+\cos ^{4} \theta=1$.
If $\displaystyle \tan \theta=\frac{1}{\sqrt{7}}$, then show that $\displaystyle \frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{3}{4}$.
Marking-scheme solution
\[\begin{aligned}
& \sin \theta+\sin ^{2} \theta=1 \\
& \Rightarrow \sin \theta=1-\sin ^{2} \theta=\cos ^{2} \theta \\
& \therefore \cos ^{2} \theta+\cos ^{4} \theta=\cos ^{2} \theta\left(1+\cos ^{2} \theta\right) \\
& =\sin \theta(1+\sin \theta) \\
& =\sin \theta+\sin ^{2} \theta=1
\end{aligned}
\]
\[\begin{aligned}
& \sec ^{2} \theta=1+\frac{1}{7}=\frac{8}{7} \\
& \cot \theta=\sqrt{7} \Rightarrow \operatorname{cosec}^{2} \theta=1+7=8 \\
& \therefore \text { LHS }=\frac{8-\dfrac{8}{7}}{8+\dfrac{8}{7}}=\frac{48}{64} \\
& \quad=\frac{3}{4}=\text { RHS }
\end{aligned}
\]
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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.