Math, asked by Anonymous, 1 year ago

prove that tan2A=2tanA/1-tan^2A

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Answered by ihrishi

Step-by-step explanation:

$tan 2A \: = \frac{2tan A}{1 - {tan}^{2}A } \\ LHS = tan 2A \\ = tan (A \: + A) \\ = \frac{tan A + tan A }{1 -tan A \times tan A } \\ = \frac{2tan A }{1 - {tan}^{2} A} \: \\ = RHS \\ thus \: proved \:$

Answered by Abhiroop12

Answer:

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tan (2A) = sin(2A) / cos(2A)

= 2 sinA cos A / (cos^2 A - sin^2 A)

Divide numerator and denominator by cos^2 A...

= 2 (sinA / cosA) / (1-[sinA / cosA]^2)

= 2 tanA / (1-tan^2 A)

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prove that tan2A=2tanA/1-tan^2A​

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prove that tan2A=2tanA/1-tan^2A