Question

A triangle with sides $a,b,c$ has corresponding angles $A,B,C$ in A.P. Then

• A. $\frac{a+c}{\sqrt{a^2+c^2-ac}}=2\cos \frac{A-C}{2}$
• B. $2\sin \frac{A}{2}\sin \frac{C}{2}=\sin \frac{B}{2}$
• C. $a^2,b^2,c^2$ are in A.P. if $a,b,c$ are in G.P.
• D. $\tan A,\tan B,\tan C$ are in A.P. if $a,b,c$ are in G.P.

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\frac{a+c}{\sqrt{a^2+c^2-ac}}=2\cos \frac{A-C}{2}$
B $2\sin \frac{A}{2}\sin \frac{C}{2}=\sin \frac{B}{2}$
C $a^2,b^2,c^2$ are in A.P. if $a,b,c$ are in G.P.

$2B=A+C\Rightarrow B=\frac{\pi }{3}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}\Rightarrow a^2-ac+c^2=b^2\Rightarrow \frac{a+c}{\sqrt{a^2+c^2-ac}}=\frac{a+c}{b}\Rightarrow \frac{a+c}{\sqrt{a^2+c^2-ac}}=\frac{\sin A+\sin C}{\sin B}\Rightarrow \frac{a+c}{\sqrt{a^2+c^2-ac}}=\frac{2\sin \frac{A+C}{2}\cos \frac{A-C}{2}}{\sin B}\Rightarrow \frac{a+c}{\sqrt{a^2+c^2-ac}}=2\cos \frac{A-C}{2}$



$2\sin \frac{A}{2}\sin \frac{C}{2}=2\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-b)(s-a)}{ba}}$
$=\frac{2(s-b)}{b}\sqrt{\frac{(s-a)(s-c)}{ac}}$
$=\frac{a+c-b}{b}\sin \frac{B}{2}=\sin \frac{B}{2}$



$a,b,c$ are in G.P.
$\Rightarrow ac=b^2$
Also, $\cos B=\frac{1}{2}=\frac{c^2+a^2-b^2}{2ca}$
$\Rightarrow 2b^2=c^2+a^2$
$\therefore a^2,b^2,c^2$ are in A.P.


$a,b,c$ are in G.P.
$\Rightarrow a^2,b^2,c^2$ are in A.P.
$\Rightarrow b^2-a^2=c^2-b^2\Rightarrow {\sin }^{2}B-{\sin }^{2}A={\sin }^{2}C-{\sin }^{2}B\Rightarrow {\sin }^2(A+C)-{\sin }^{2}A={\sin }^{2}C-{\sin }^2(A+C)\Rightarrow \sin \left(C\right)\sin (2A+C)=-\sin \left(A\right)\sin (2C+A)\Rightarrow \sin \left(C\right)\sin (\pi -(B-A\left)\right)=-\sin \left(A\right)\sin (\pi -(B-C\left)\right)\Rightarrow \sin \left(C\right)\left[\sin \right(B\left)\cos \right(A)-\cos (B\left)\sin \right(A\left)\right]=-\sin \left(A\right)\left[\sin \right(B\left)\cos \right(C)-\cos (B\left)\sin \right(C\left)\right]$

$\Rightarrow \sin C(\sin B\cos A-\cos B\sin A)=\sin A(\sin C\cos B-\cos C\sin B)$

Dividing both the sides by $\sin A\sin B\sin C$, we get
$\cot A,\cot B,\cot C$ are in A.P.
$\Rightarrow \tan A,\tan B,\tan C$ are in H.P.