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,

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

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

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

$=costheta/(sin2thetacostheta)$

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

$=color(blue)(csc2theta$