Triangle abc is not a right triangle and cos(a)=cos(b)cos(c) show that tan(b)tan

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cos(a)=cos(b)cos(c) we know that in an triangle the sum of angles should be 180 degrees so since a,b,c are angles of a triangle we know that a=180-(b+c) a+b+c=180 =>a=180-(b+c) cos(180-(b+c))=cosb cosc we know that cos(180-x)=-cosx -cos (b+c) = cos b cos c multiply by -1 on both sides we get cos(b+c)=-cosb cosc we know that cos(b+c)=cosb cosc -sinb sinc cosb cosc-sinb sinc=-cosb*cosc =>cosb cosc+cos b cosc= sinb*sinc =>2 cosb cos c=sinb sinc dividing by cosb * cosc on both sides since b and c are not 90 so dividing by them is correct =>(sinb/cosb)*(sinc/cosc)=2 we know that tanx =sinx/cosx =>tanb tanc =2 hence showed

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