Sin60=2tan 30/1+ tan²30, Verify It.
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Solve this as shown in the attachment.
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sin60° = √3/2
(2tan30°)/(1 + tan²30°)
we know, tan30° = 1/√3
so, (2tan30°)/(1 + tan²30°) = (2 × 1/√3)/{1 + (1/√3)²}
= (2/√3)/(1 + 1/3)
= (2/√3)/(4/3)
= 2/√3 × 3/4
= 2/√3 × (√3)²/2²
= √3/2
here it is clear that sin60° = (2tan30°)/(1 + tan²30°)
hence verified
(2tan30°)/(1 + tan²30°)
we know, tan30° = 1/√3
so, (2tan30°)/(1 + tan²30°) = (2 × 1/√3)/{1 + (1/√3)²}
= (2/√3)/(1 + 1/3)
= (2/√3)/(4/3)
= 2/√3 × 3/4
= 2/√3 × (√3)²/2²
= √3/2
here it is clear that sin60° = (2tan30°)/(1 + tan²30°)
hence verified
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