Question
If $$\cos A + \cos B + 2\cos C = 2$$ then the sides of the $$\vartriangle ABC$$ are in
A.
A.P.
B.
G.P.
C.
H.P.
D.
None of these
Answer :
A.P.
Solution :
$$\eqalign{
& 2\cos \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2} = 2{\sin ^2}\frac{C}{2} \cr
& \Rightarrow \,\,\cos \frac{{A - B}}{2} = 2\cos \frac{{A + B}}{2}\,\,\,\,\left( {\because \,\,\cos \frac{{A + B}}{2} \ne 0} \right) \cr
& {\text{or, }}\cos \frac{A}{2} \cdot \cos \frac{B}{2} + \sin \frac{A}{2} \cdot \sin \frac{B}{2} = 2\,\,\left( {\cos \frac{A}{2} \cdot \cos \frac{B}{2} - \sin \frac{A}{2} \cdot \sin \frac{B}{2}} \right) \cr
& {\text{or, }}3\sin \frac{A}{2} \cdot \sin \frac{B}{2} = \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cr
& \Rightarrow \,\,\tan \frac{A}{2} \cdot \tan \frac{B}{2} = \frac{1}{3} \cr
& {\text{or, }}\sqrt {\frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}} \cdot \sqrt {\frac{{\left( {s - c} \right)\left( {s - a} \right)}}{{s\left( {s - b} \right)}}} = \frac{1}{3} \cr
& \Rightarrow \,\,\frac{{s - c}}{s} = \frac{1}{3} \cr
& \Rightarrow \,\,\frac{{a + b - c}}{{a + b + c}} = \frac{1}{3} \cr
& \Rightarrow \,\,3a + 3b - 3c = a + b + c \cr
& \Rightarrow \,\,a + b = 2c \cr
& \Rightarrow \,\,a,b,c\,\,{\text{are in A}}{\text{.P}}{\text{.}} \cr} $$