Question #41db9

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Answer 1 · 2016-10-05T23:40:15

${sec}^{2} θ + 1 - 3 tan θ = 0$ $sec^2theta+1-3tantheta=0$

$\Rightarrow 1 + {tan}^{2} θ + 1 - 3 tan θ = 0$ $=>1+tan^2theta+1-3tantheta=0$

$\Rightarrow {tan}^{2} θ - 3 tan θ + 2 = 0$ $=>tan^2theta-3tantheta+2=0$

$\Rightarrow {tan}^{2} θ - 2 tan θ - tan θ + 2 = 0$ $=>tan^2theta-2tantheta-tantheta+2=0$

$\Rightarrow (tan θ - 2) tan θ - 1 (tan θ - 2) = 0$ $=>(tantheta-2)tantheta-1(tantheta-2)=0$

$\Rightarrow (tan θ - 2) (tan θ - 1) = 0$ $=>(tantheta-2)(tantheta-1)=0$

So

when $tan θ = 2 = {tan tan}^{- 1} (2) = tan α$ $tantheta=2=tantan^-1(2)=tanalpha$

$=>theta=npi+alpha." where "n inZZ$

when $tantheta=1=tan(pi/4)$

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