sin 2A ≡ (2 tan A)/(1 +tan² A)
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Answered by
53
ANSWER :
We need to prove that:
sin2A = 2tanA/ (1+tan^2 A)
We will start from the right side and prove the left side.
We know that tanA = sinA/cosA
OR..........
==> 2tanA / (1+tan^2 A) = 2(sinA/cosA) / [1+ (sinA/cosA)^2]
= 2sinA/ cosA[ 1+ (sin^2A/cos^2 A)]
= 2sinA/ cosA*[(cos^2 A + sin^2 A)/cos^2A]
But we know that sin^2 A + cos^2 A = 1
==> 2tanA/ (1+tan^2 A) = 2sinA/ (1/cosA)
= 2sinA*
You want the following to be proved: sin 2A = 2tan A/ (1 + (tan A)^2).
We know that tan A = sin A/ cos A and (sin A)^2 + (cos A)^2 = 1.
2tan A/ (1 + (tan A)^2)
=> [2* sin A / cos A]/ [ 1 + (sin A)^2 / (cos A)^2]
=> [2* sin A / cos A]/ [ ((cos A)^2 + (sin A)^2) / (cos A)^2]
=> [2 sin A* (cos A)^2 / cos A]/ [ (cos A)^2 + (sin A)^2]
=> [2 sin A * cos A] / 1
=> sin 2A
Therefore sin 2A = 2tan A/ (1 + (tan A)^2).
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PLZ MARK THE ANSWER AS BRAINLY ANSWER
We need to prove that:
sin2A = 2tanA/ (1+tan^2 A)
We will start from the right side and prove the left side.
We know that tanA = sinA/cosA
OR..........
==> 2tanA / (1+tan^2 A) = 2(sinA/cosA) / [1+ (sinA/cosA)^2]
= 2sinA/ cosA[ 1+ (sin^2A/cos^2 A)]
= 2sinA/ cosA*[(cos^2 A + sin^2 A)/cos^2A]
But we know that sin^2 A + cos^2 A = 1
==> 2tanA/ (1+tan^2 A) = 2sinA/ (1/cosA)
= 2sinA*
You want the following to be proved: sin 2A = 2tan A/ (1 + (tan A)^2).
We know that tan A = sin A/ cos A and (sin A)^2 + (cos A)^2 = 1.
2tan A/ (1 + (tan A)^2)
=> [2* sin A / cos A]/ [ 1 + (sin A)^2 / (cos A)^2]
=> [2* sin A / cos A]/ [ ((cos A)^2 + (sin A)^2) / (cos A)^2]
=> [2 sin A* (cos A)^2 / cos A]/ [ (cos A)^2 + (sin A)^2]
=> [2 sin A * cos A] / 1
=> sin 2A
Therefore sin 2A = 2tan A/ (1 + (tan A)^2).
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PLZ MARK THE ANSWER AS BRAINLY ANSWER
Answered by
5
Answer:
We know that tan A = sin A/ cos A and (sin A)^2 + (cos A)^2 = 1.
2tan A/ (1 + (tan A)^2)
=> [2* sin A / cos A]/ [ 1 + (sin A)^2 / (cos A)^2]
=> [2* sin A / cos A]/ [ ((cos A)^2 + (sin A)^2) / (cos A)^2]
=> [2 sin A* (cos A)^2 / cos A]/ [ (cos A)^2 + (sin A)^2]
=> [2 sin A * cos A] / 1
=> sin 2A
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