question_answer

               

Show that \[\Delta ABC\] is an isosceles triangle, if the determinant

OR

If \[A+B+C=\pi ,\] show that

\[=-\sin (A-B)\sin (B-C)\sin (C-A)\]

Answer:

We have,

Applying \[{{C}_{1}}\to {{C}_{1}}-{{C}_{3}}\] and \[{{C}_{2}}\to {{C}_{2}}-{{C}_{3}},\] we get

Taking \[(\cos A-\cos C)\]common from \[{{C}_{1}}\] and \[(\cos B-\cos C)\] common from \[{{C}_{2}},\] we get

\[(\cos A-\cos C)\cdot (\cos B-\cos C)\]

\[\left. \begin{align}   & 1 \\  & 1+\cos C \\  & {{\cos }^{2}}C+\cos C \\ \end{align} \right]=0\]

\[\Rightarrow \,\,\,\,(\cos A-\cos C)\cdot (\cos B-\cos C)\]

\[[(\cos B+\cos C+1)-(\cos A+\cos C+1)]=0\]

[Expanding along \[{{R}_{1}}\]]

\[\Rightarrow \,\,\,\,\,(\cos A-\cos C)\cdot (\cos B-\cos C)\]

\[(\cos B+\cos C+1-\cos A-\cos C-1)=0\]

\[\Rightarrow \,\,(\cos A-\cos C)\cdot (\cos B-\cos C)(\cos B-\cos A)=0\]\[\Rightarrow \] \[\cos A=\cos C\,\,\text{or}\,\,\cos B=\cos C\,\,\text{or}\,\,\cos B\]

\[=\cos A\]

\[\Rightarrow \,\,\,\,A=C\text{ or }B=C\text{ or }B=A\]

Hence, ABC is an isosceles triangle.

Hence proved.

OR

The given determinant

\[[{{C}_{1}}\to {{C}_{1}}+{{C}_{3}}]\]

[taking \[2sin(A-B)\] and \[2sin(A\,-C)\] common from \[{{R}_{2}}\] and \[{{R}_{3}}\]respectively]

\[[\because \,\,\,A+B+C=\pi ]\]

\[=\sin (A-B)\sin (A-C)[\sin B\cos C-\cos B\sin C]\]

\[=\sin (A-B)\sin (A-C)\sin (B-C)\]

\[=-\sin (A-B)\sin (B-C)\sin (C-A)\]