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If sin2a =#sin2b, then value of tan(a+ b) /tan(a-b)?
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If sin2a =#sin2b, then value of tan(a+ b) /tan(a-b)?
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If sin2a =#sin2b, then value of tan(a+ b) /tan(a-b)?
1. Given:
- sin2a = sin2b
2. Key Point:
- sin2a = 2sinacos a
- sin2b = 2sinbcos b
3. Calculation:
- 2sinacos a = 2sinbcos b
- sin(a)cos(a) = sin(b)cos(b)
- tan(a) = tan(b)
4. Formula for tan(a + b):
- tan(a + b) = (tan a + tan b) / (1 - tan a * tan b)
5. Formula for tan(a - b):
- tan(a - b) = (tan a - tan b) / (1 + tan a * tan b)
6. Result:
- tan(a + b) / tan(a - b) = (tan a + tan b) / (tan a - tan b) = 1
Therefore, if sin2a = sin2b, then the value of tan(a + b) / tan(a - b) is equal to 1.
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If sin2a =#sin2b, then value of tan(a+ b) /tan(a-b)?
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