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Question Number 43642 by gunawan last updated on 13/Sep/18
The most general value of θ which  satisfy both the equations cos θ= −(1/( (√2)))  and  tan θ=1  is
Themostgeneralvalueofθwhichsatisfyboththeequationscosθ=12andtanθ=1is
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Sep/18
cosθ=−ve   tanθ=+ve   so θ lies in third quadrant  tanθ=1=tan(Π+(Π/4))  θ=((5Π)/4)  check...  cos(Π+(Π/4))=−cos((Π/4))=−(1/( (√2)))  tanθ=tan((5Π)/4)  θ=xΠ+((5Π)/4)  x=2  θ=((13Π)/4)    cosθ=cos(((5Π)/4))  θ=2yΠ±((5Π)/4)  θ=2yΠ+((5Π)/4)  y=1   θ=((13Π)/4)
cosθ=vetanθ=+vesoθliesinthirdquadranttanθ=1=tan(Π+Π4)θ=5Π4checkcos(Π+Π4)=cos(Π4)=12tanθ=tan5Π4θ=xΠ+5Π4x=2θ=13Π4cosθ=cos(5Π4)θ=2yΠ±5Π4θ=2yΠ+5Π4y=1θ=13Π4

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