Question

prove that 1+tan2AtanA=sec2A​

Answer 1

Step-by-step explanation:

1+tan2A/tanA

sec2A//sinA/cosA

1/cos2A

sec2A

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Answer 2

Answer:

$Proved !!$

Step-by-step explanation:

$1 + \tan\,2 A (\tan\,A) = \sec\,2 A\text{Right hand side}\sec\,2 A = \dfrac{1}{\cos\,2 A} =\dfrac{1}{\cos^2\,A - \sin^2\,A}=\dfrac{cos^2\,A + \sin^2\,A}{\cos^2\,A - \sin^2\,A}=\dfrac{1 + \tan^2\,A}{1 - \tan^2\,A}= \dfrac{1 - \tan^2\,A + 2\tan^2\,A}{1 - \tan^2\,A}= \dfrac{1 - \tan^2\,A}{1 - \tan^2\,A} + \dfrac{2\tan^2\,A}{1 - \tan^2\,A}= 1 + \dfrac{2\tan\,A}{1 - \tan^2\,A}\times \tan\,A$

$=1+ tan2AtanA$