Question

If the tangents of the angles of a triangle are in A.P. prove that the squares of the sides are in the ratio $x^2(x^2+9):(3+x^2{)}^2:9(1+x^2)$, where x is tangent of the least or greatest angle.

Answer 1

Since tan A , tan B, tan C are in A.P., we have
$tanA+tanC=2tanB$ ....(1)
or $x+tanC=2tanB$ ......(2)
Where $x=tanA.$
[Observe that if tan A, tan B, tan C are in A.P. then tan A is either the greatest of the least amongst tan A , tan B and tan C].
Now in a triangle ABC, we always have
tan A tan B tan C = tan A + tan B + tan C .......(3)
$\therefore$ From (1), (2) and (3), we obtain
x tan B(2tan B -x)= 2 tan B+ tan B
or x(2tan B- x)=3 [$\because tanB\ne 0$]
$\therefore tanB=(3+x^2)/2x.$
and $tanC=2tanB-x=(3+x^2)/x-x=3/x.$
Now in a$△ABC$ , we have
$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$
$\Rightarrow \frac{a^2}{sin62A}=\frac{b^2}{si{n}^{2}B}=\frac{c^2}{si{n}^{2}C}$
Hence $a^2:b^2:c^2=si{n}^{2}A:si{n}^{2}B:si{n}^{2}C$
But $tanA=x\Rightarrow sinA=x\sqrt{1+x^2}$
$\Rightarrow si{n}^{2}A=x^2/(1+x^2)$
and$tanB=(3+x^2)/2x$
$\Rightarrow si{n}^{2}B=(3+x^2{)}^2/[(3+x^2{)}^2+4x^2]$
$=(3+x^2{)}^2/(1+x^2\left)\right(9+x^2)$
and $tanC=3/x\Rightarrow sinC=3/\sqrt{(9+x^2)}$
$\Rightarrow si{n}^{2}C=9/(9+x^2)$
$\therefore a^2:b^2:c^2::\frac{x^2}{1+x^2}:\frac{(3+x^2{)}^2}{(1+x^2)(9+x^2)}:\frac{9}{x^2+9}::x^2(9+x^2):(3+x^2{)}^2:9(1+x^2).$
Ambiguous case: When two sides say a and b are given and one angle opposite to these sides say A is given, then by sine rule $\frac{a}{sinA}=\frac{b}{sinB}$
$\therefore sinB=\frac{b}{a}sinA=k,$ say.
There will be two values of angle B say $B_1$ and $B_2$ which will satisfy the equation sin B =k.
Evidently these two values are supplementary i.e. if $B_1=\alpha$ then $B_2=\pi -\alpha$ i.e. $B_1+B_2=\pi$