If tan2Atan4A = 1 , find tan3A = ?
rohitkumargupta:
Heya its power or multiplication
Tan²A ×3tan⁴A or tan2A ×3
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tan(x - y) = (tan x - tan y) / (1 + tan x tan y) , so
tan(4A - 2A) = (tan 4A - tan 2A) / (1 + tan 4A * tan 2A)
tan 2A = (tan 4A - tan 2A) / 2 ; as tan 4A * tan 2A = 1
2 tan (2A) = tan (4A) - tan (2A)
3 tan (2A) = tan (4A)
plug this value to original equation tan (2A) * tan (4A) = 1, we get
tan (2A) * 3 tan (2A) = 1
tan² (2A) = 1/3
tan (2A) = 1/√3
2A = 30°
=>A = 15°
therefore 3A = 45°, and tan (3A) = tan (45°) = 1
tan(4A - 2A) = (tan 4A - tan 2A) / (1 + tan 4A * tan 2A)
tan 2A = (tan 4A - tan 2A) / 2 ; as tan 4A * tan 2A = 1
2 tan (2A) = tan (4A) - tan (2A)
3 tan (2A) = tan (4A)
plug this value to original equation tan (2A) * tan (4A) = 1, we get
tan (2A) * 3 tan (2A) = 1
tan² (2A) = 1/3
tan (2A) = 1/√3
2A = 30°
=>A = 15°
therefore 3A = 45°, and tan (3A) = tan (45°) = 1
Answered by
1
Step-by-step explanation:
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tan(x - y) = (tan x - tan y) / (1 + tan x tan y) , so
tan(4A - 2A) = (tan 4A - tan 2A) / (1 + tan 4A * tan 2A)
tan 2A = (tan 4A - tan 2A) / 2 ; as tan 4A * tan 2A = 1
2 tan (2A) = tan (4A) - tan (2A)
3 tan (2A) = tan (4A)
plug this value to original equation tan (2A) * tan (4A) = 1, we get
tan (2A) * 3 tan (2A) = 1
tan² (2A) = 1/3
tan (2A) = 1/√3
2A = 30°
=>A = 15°
therefore 3A = 45°, and tan (3A) = tan (45°) = 1
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