Cot2θ+tanθ = ? Write Answer With Explanation...

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Answer 1 · 2018-05-02T06:06:56

#=csc2theta#

We know that,

#(1)tan2x=(2tanx)/(1-tan^2x)#

#(2)sin2x=(2tanx)/(1+tan^2x)#

Using #(1)and(2)#

#cot2theta+tantheta=1/tan(2theta)+tantheta#

#=1/((2tantheta)/(1-tan^2theta))+tantheta#

#=(1-tan^2theta)/(2tantheta)+tantheta#

#=(1-tan^2theta+2tan^2theta)/(2tantheta)#

#=(1+tan^2theta)/(2tantheta)#

#=1/((2tantheta)/(1+tan^2theta))#

#=1/(sin2theta)#

#=csc2theta#

Answer 2 · 2018-05-02T06:48:23

#=csc2theta#

We know that,

#color(red)((1)cosAcosB+sinAsinB=cos(A-B)#

#color(blue)((2)cscA=1/sinA#

Now,

#cot2theta+tantheta=(cos2theta)/(sin2theta)+sintheta/costheta#

#=color(red)((cos2thetacostheta+sin2thetasintheta))/(sin2thetacostheta)..tocolor(red)(Apply(1)#

#=color(red)((cos(2theta-theta)))/(sin2thetacostheta)#

#=costheta/(sin2thetacostheta)#

#=color(blue)(1/(sin2theta)...toApply(2)#

#=color(blue)(csc2theta#