Answer:
[tex]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) \times 2 }{1 - {( \tan(x) }^{4} } \\ = \frac{4 \tan(x) }{1 - {( \tan( \times ) )}^{4} } [/tex]
Step-by-step explanation:
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