Trigonometry problem.

C=180-(A+B) so cosC=-cos(A+B), sinC=sin(A+B) and tanC=-tan(A+B).

But cosA=cosBcosC=-cosBcos(A+B), so sinA=sqrt(1-cos^2A)=sqrt(1-cos^Bcos^2(A+B)).

cos(A+B)=cosAcosB-sinAsinB, sin(A+B)=sinAcosB+cosAsinB, tan(A+B)=(tanA+tanB)/(1-tanAtanB); cos2B=2cos^2B-1, cos^2B=(cos2B+1)/2; sin2B=2sinBcosB (standard trigonometric identities).

Let cosA=a and cosB=b, so a=-b(ab-sqrt((1-a^2)(1-b^2)))=-ab^2+bsqrt((1-a^2)(1-b^2)).

a(1+b^2)=bsqrt((1-a^2)(1-b^2)); squaring both sides: a^2(1+b^2)^2=b^2(1-a^2)(1-b^2);

)a^2+2a^2b^2+a^2b^4=b^2(1-a^2-b^2+a^2b^2)=b^2-a^2b^2-b^4+a^2b^4;

a^2+3a^2b^2=b^2-b^4; a^2=b^2(1-b^2)/(1+3b^2) and a=bsqrt((1-b^2)/(1+3b^2)), so we have a in terms of b. We can also write: cosA=cosBsinB/sqrt(1+3cos^2B)=sin2B/2sqrt(1+(3/2)(cos2B+1))=sin2B/sqrt(10+6cos2B).

From this sinA=sqrt(1-sin^2(2B)/(10+6cos2B))=sqrt((10+6cos2B-sin^2(2B))/(10+6cos2B))

=sqrt((10+6cos2B-1+cos^2(2B))/(10+6cos2B))=sqrt((9+6cos2B+cos^2(2B))/(10+6cos2B))=

(3+cos2B)/sqrt(10+6cos2B), and tanA=sinA/cosA=(3+cos2B)/sin2B=3cosec2B+cot2B.

Also, tanA=sinA/cosA=sqrt(1-a^2)/a; tanB=sqrt(1-b^2)/b; tanC=(sqrt(1-a^2)/a+sqrt(1-b^2)/b)/(1-sqrt((1-a^2)(1-b^2))/ab)=(bsqrt(1-a^2)+asqrt(1-b^2))/(ab-sqrt((1-a^2)(1-b^2))).

Or, tanC=(3cosec2B+cot2B+tanB)/(1-(3cosec2B+cot2B)tanB).

tanA-tanB-tanC=(bsqrt(1-a^2)-asqrt(1-b^2))/ab-(bsqrt(1-a^2)+asqrt(1-b^2))/(ab-sqrt((1-a^2)(1-b^2)))

=tanA-tanB+tan(A+B)=tanA-tanB+(tanA+tanB)/(1-tanAtanB)

=(tanA-tanB-(tanA-tanB)tanAtanB+tanA+tanB)/(1-tanAtanB)=(2tanA-(tanA-tanB)tanAtanB)/(1-tanAtanB)

=tanA(2-tanB(tanA-tanB)tanB)/(1-tanAtanB).

tanA-tanB-tanC=3cosec2B+cot2B-tanB-(3cosec2B+cot2B+tanB)/(1-tanB(3cosec2B+cot2B))=

(-2tanB-tanB(3cosec2B+cot2B)(3cosec2B+cot2B-tanB))/(1-tanB(3cosec2B+cot2B).

[This is the same as:

sin(A-B)/cosAcosB-sin(A+B)/cos(A+B)=(sin(A-B)cos(A+B)-sin(A+B)cosAcosB)/(cosAcosBcos(A+B)).

The question doesn't specify in what terms tanA-tanB-tanC is to be expressed. tanA and tanB are mutually dependent values, but tanC depends on angles A and B. For example, tanB and tanC can both be expressed in terms of A, tanA and tanC can both be expressed in terms of B, etc. Similarly, tanA-tanB-tanC can be expressed in A, or B, or C.]

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Prouve TanB=(b×sinC)/(a-b×cos). B and C are the angles a and b the distances
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