Question

Tan 2A + sin 2A=4 tan A/(1 - tan4 A)​

Answer 1

`when \: \\ \tan(2x) = \frac{2 \tan(x) }{1 - {( \tan(x)) }^(2) } \: \: and \\ \sin(2x) = \frac{2 \tan(x) }{1 + {( \tan(x)) }^(2) } \\ then \\ \sin(2x) + \tan(2x) = \frac{2 \tan(x) }{1 - {( \tan(x)) }^(2) } + \frac{2 \tan(x) }{1 + {( \tan(x)) }^(2) } \\ = \frac{2 \tan(x)(1 + {( \tan(x)) }^(2) + 2 \tan(x) (1 - {( \tan(x)) }^(2) }{(1 + {( \ \tan(x)) }^(2)(1 - {( \tan(x)) }^(2) } \\ = \frac{2 \tan(x) * 2 }{1 - {( \tan(x) }^(4) } \\ = \frac{4 \tan(x) }{1 - {( \tan( * ) )}^(4) }`

Explanation:

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