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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 4 · Set 2 · Q21
Evaluate : $\displaystyle \frac{5}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cot ^{2} 45^{\circ}+2 \sin ^{2} 90^{\circ}$If $\displaystyle \theta$ is an acute angle and $\displaystyle \sin \theta=\cos \theta$, find the value of $\displaystyle \tan ^{2} \theta+\cot ^{2} \theta-2$.
Evaluate : $\displaystyle \frac{5}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cot ^{2} 45^{\circ}+2 \sin ^{2} 90^{\circ}$
If $\displaystyle \theta$ is an acute angle and $\displaystyle \sin \theta=\cos \theta$, find the value of $\displaystyle \tan ^{2} \theta+\cot ^{2} \theta-2$.
Marking-scheme solution
\[\begin{aligned}
& \frac{5}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cot ^{2} 45^{\circ}+2 \sin ^{2} 90^{\circ} \\
& =\frac{5}{(\sqrt{3})^{2}}+\frac{1}{(\sqrt{3} / 2)^{2}}-(1)^{2}+2(1)^{2}=\frac{5}{3}+\frac{4}{3}-1+2 \\
& =\frac{9}{3}+1=4
\end{aligned}
\]
\[\begin{aligned}
& \sin \theta=\cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=1 \Rightarrow \tan \theta=1 \Rightarrow \cot \theta=1 \\
& \tan ^{2} \theta+\cot ^{2} \theta-2=(1)^{2}+(1)^{2}-2=0
\end{aligned}
\]
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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.