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#.