Solve the trigonometric equation #tantheta+cottheta=2#?

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Answer 1 · 2017-10-08T15:23:55

#theta=npi+pi/4#

#tantheta+cottheta=2#

i.e. #sintheta/costheta+costheta/sintheta=2#

or #(sin^2theta+cos^2theta)/(sinthetacostheta)=2#

or #1/(sinthetacostheta)=2#

or #2sinthetacostheta=1#

or #sin2theta=sin(pi/2)#

Hence #2theta=2npi+pi/2#

or #theta=npi+pi/4#

Answer 2 · 2017-10-08T15:26:18

#tantheta+cottheta=2#

#=>tantheta+1/tantheta=2#

#=>tan^2theta-2tantheta+1=0#

#=>(tantheta-1)^2=0#

#=>tantheta=1=tan(pi/4)#

#=>theta=npi+pi/4" where "n in ZZ#