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CBSE Class 10 Standard Math 2025 All Sets Solved Paper

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Question : 3 of 20
Marks: +1, -0
The value of (1−2sin⁡260∘)(1 - 2\sin^2 60^{\circ}) is same as that of
Solution:  
1−2sin⁡260∘1-2\sin^2 60^{\circ}
=1−2×(32)2=1-2\times \left(\frac{\sqrt{3}}{2}\right)^2
=1−2×34=1-2\times \frac{3}{4}
=1−32=1-\frac{3}{2}
=2−32=\frac{2-3}{2}
=−12=-\frac{1}{2}
=−sin⁡230∘=-\sin^2 30^{\circ}
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