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CBSE Class 10 Basic Math 2026 All Sets Solved Paper

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Question : 16 of 20
Marks: +1, -0
The value of 2tan⁡45∘sin⁡260∘−cos⁡90∘2\tan 45^{\circ}\sin^2 60^{\circ} - \cos 90^{\circ} is:
Solution:  
2 tan⁡ 45∘sin⁡2 60∘−cos⁡ 90∘2\,\tan\,45^{\circ}\sin^2\,60^{\circ} - \cos\,90^{\circ}
=2×1×(  32)2−0= 2 \times 1 \times \left(\; \frac{\sqrt{3}}{2}\right)^2 - 0
=2×  34= 2 \times \; \frac{3}{4}
=  32= \; \frac{3}{2}
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