Prove: sec*4x-tan*4x=1+2tan*2x

(sin/sin+1)-(sin/sin-1)=-2tan^2

sin2 =2tan cos

sin^2x+sin^2tan^2x=tan^2x

sec^4x-tan^4x=1+2tan^2x

cos(2tan^-1(15/8))

cot^-1(2tan(cos^-18/17))+tan^-1(2tan(sin^-18/17))=tan^-1300/161

2tan^2x-tanx-1=0 0<x<2pi

2tan^2x-tanx-1=0 0<x<2pi •MATH - Steve, Thursday, July 18, 2013 at 11:48am (2tanx+1)(tanx-1)...

2tan^-1(tanx/2×tan(π/4-y/2))=tan^-1(sinx×siny/cosx+cosy)

10. Rewrite 2tan 23° / 1 - tan223° as a single function of an angle. I'd appreciate it if someone...

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