How do you simplify #(1+tantheta)/(1+cottheta)#?

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Answer 1 · 2018-08-01T04:56:11

#(1+tantheta)/(1+cottheta)=tantheta#

Here,

#(1+tantheta)/(1+cottheta)=(1+tantheta)/(1+1/tantheta)to[becausecottheta=1/tantheta]#

#=>(1+tantheta)/(1+cottheta)= (1+tantheta)/((tantheta+1)/tantheta)#

#=>(1+tantheta)/(1+cottheta)=tantheta((1+tantheta)/(tantheta+1))#

#:.(1+tantheta)/(1+cottheta)=tantheta#

Answer 2 · 2018-08-01T06:04:03

#color(purple)(=> sin theta / cos theta = tan theta#

#(1 + tan theta) / (1 + cot theta)#

#=> (1 + sin theta/cos theta) / (1 + cos theta / sin theta)#

#=> (cancel(cos theta + sin theta)/cos theta) / (cancel(sin theta + cos theta)/sin theta)#

#color(purple)(=> sin theta / cos theta = tan theta#