CBSE 2025 · Region 1 · Set 1 · Q21 · 2 marks
If $\displaystyle \mathrm{x} \cos 60^{\circ}+\mathrm{y} \cos 0^{\circ}+\sin 30^{\circ}-\cot 45^{\circ}=5$, then find the value of $\displaystyle \mathrm{x}+2 \mathrm{y}$.Evaluate : $\displaystyle \frac{\tan ^{2} 60^{\circ}}{\sin ^{2} 60^{\circ}+\cos ^{2} 30^{\circ}}$
If $\displaystyle \mathrm{x} \cos 60^{\circ}+\mathrm{y} \cos 0^{\circ}+\sin 30^{\circ}-\cot 45^{\circ}=5$, then find the value of $\displaystyle \mathrm{x}+2 \mathrm{y}$.
Evaluate : $\displaystyle \frac{\tan ^{2} 60^{\circ}}{\sin ^{2} 60^{\circ}+\cos ^{2} 30^{\circ}}$
Marking-scheme solution
\[\begin{aligned}
& \mathrm{x}\left(\frac{1}{2}\right)+\mathrm{y}(1)+\frac{1}{2}-1=5 \\
\Rightarrow & \mathrm{x}+2 \mathrm{y}=11
\end{aligned}
\]
\[\frac{(\sqrt{3})^{2}}{\left(\dfrac{\sqrt{3}}{2}\right)^{2}+\left(\dfrac{\sqrt{3}}{2}\right)^{2}}
\]
Introduction to TrigonometryTrigonometric Ratios of Some Specific AnglesApplyvery_short_answereasy
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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.