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September 19, 2014

September 19, 2014

Posted by **anonymous** on Friday, June 18, 2010 at 10:40pm.

[sinx + tanx] / [cosx +1] = tanx

- maths - trigonometry -
**Ariel**, Friday, June 18, 2010 at 10:50pmI just got done with thi stuff and these are the formulas that helped me with alot.

tan(angle)=opposite/adjacent

coa(angle)=adjacent/hypotenuse

sin(angle)=opposite/hypotenuse

Is this what your asking for?

- maths - trigonometry -
**Reiny**, Saturday, June 19, 2010 at 12:13amthat's sort of a tricky one,

[sinx + tanx] / [cosx +1] = tanx

LS = (sinx + tanx)/(cosx+1) [(cosx-1)/cosx-1)]

= (sinxcosx - sinx + sinx - sinx/cosx)/(cos^2x - 1)

= [ (sinxcos^2x - sinx)/cosx]/(cos^2x - 1)

= sinx[cos^2x - 1)/cosx] / (cos^2x - 1)

= sinx/cosx

= tanx

= RS

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