Tanx/1-cotx+cotx/1-tanx=?

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Answer 1 · 2018-03-29T09:39:06

$tanx/(1-cotx)+ cotx/(1-tanx)$

$=(tanx(-tanx))/((1-cotx)(-tanx))+ cotx/(1-tanx)$

$= (- tan^2x)/(1-tanx) +cotx/(1-tanx)$ $[because" tanx xx cotx = 1]$

$=(cotx-tan^2x)/(1-tanx)$

$=(1/tanx-tan^2x)/(1-tanx)$

$=(1-tan^3x)/(tanx(1-tanx))$

$=((1-tanx)(1+tanx+tan^2x))/(tanx(1-tanx))$

$=((1+tanx+tan^2x))/(tanx)$

$=1/tanx+tanx/tanx+tan^2x/tanx)$

$=color(red)(1+tanx+cotx$

$=1+ sinx/cosx + cosx/sinx$

$=(sinxcosx + sin^2x + cos^2x)/ (sinxcosx)$

$=color(red)((sinxcosx + 1)/ (sinxcosx)$

$= 1 + 1/(sinxcosx)$

$= 1 + 2/(2sinxcosx)$

$=color(red)( 1 + 2/(sin2x)$

The final answer may vary, depending on where you may wanna stop :)

Answer 2 · 2018-03-29T09:41:24

$tanx/(1-cotx)+cotx/(1-tanx)=secxcscx+1$ .

$tanx/(1-cotx)+cotx/(1-tanx)$ ,

$=(sinx/cosx)/(1-cosx/sinx)+(cosx/sinx)/(1-sinx/cosx)$ ,

$={sinx/cosx-:(sinx-cosx)/sinx}+{cosx/sinx-:(cosx-sinx)/cosx}$ ,

$={sinx/cosx xx sinx/(sinx-cosx)}+{cosx/sinx xx cosx/(cosx-sinx)}$ ,

$=sin^2x/{cosx(sinx-cosx)}-cos^2x/{sinx(sinx-cosx)}$ ,

$=1/(sinx-cosx){sin^2x/cosx-cos^2x/sinx}$ ,

$=1/(sinx-cosx){(sin^3x-cos^3x)/(sinxcosx)}$ ,

$=1/cancel((sinx-cosx)){cancel((sinx-cosx))(sin^2x+sinxcosx+cos^2x)}/(sinxcosx)$ ,

$={(sin^2x+cos^2x)+sinxcosx}/(sinxcosx)$ ,

$=(1+sinxcosx)/(sinxcosx)$ ,

$=1/(sinxcosx)+(sinxcosx)/(sinxcosx)$ ,

$=1/sinx*1/cosx+1$ ,

$rArr tanx/(1-cotx)+cotx/(1-tanx)=secxcscx+1$ .