Question

If $A+B+C=\pi $ , prove that $\left| \begin{matrix}
   {{\sin }^{2}}A & \cot A & 1 \\
   {{\sin }^{2}}B & \cot B & 1 \\
   {{\sin }^{2}}C & \cot C & 1 \\
\end{matrix} \right|=\text{0}$

Answer 1

Hint: To solve this question we will first using elementary row operation ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$ and ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$ then we will us trigonometric identities which are $\cot x=\dfrac{\cos x}{\sin x}$, $\sin A+\operatorname{sinB}=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$, $\sin A-\operatorname{sinB}=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$ and algebraic identity ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$to solve the determinant. At last we will expand determinant along column ${{C}_{3}}$ to obtain a solution.

Complete step-by-step answer:
Now, before we start solving the questions, let us see how we calculate determinant and what are its various properties
Now , if we want to calculate the determinant of matrix A of order $3\times 3$, then determinant of matrix A of $3\times 3$ is evaluated as,
$\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$
Some of the properties of determinant are as follows,
( a ) Determinant evaluated across any row or column is the same.
( b ) If an element of a row or a column are zeros, then the value of the determinant is equal to zero.
( c ) If rows and columns are interchanged then the value of the determinant remains the same.
( d ) Determinant of an identity matrix is 1.

Now, in question it given that, If $A+B+C=\pi $and asked to prove that $\left| \begin{matrix}
   {{\sin }^{2}}A & \cot A & 1 \\
   {{\sin }^{2}}B & \cot B & 1 \\
   {{\sin }^{2}}C & \cot C & 1 \\
\end{matrix} \right|=\text{0}$.
Now, using elementary row operation ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$, we get
\[\left| \begin{matrix}
   {{\sin }^{2}}A-{{\sin }^{2}}B & \cot A-\cot B & 0 \\
   {{\sin }^{2}}B & \cot B & 1 \\
   {{\sin }^{2}}C & \cot C & 1 \\
\end{matrix} \right|=\text{0}\]
Now, using, using elementary row operation ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$, we get

\[\left| \begin{matrix}
   {{\sin }^{2}}A-{{\sin }^{2}}B & \cot A-\cot B & 0 \\
   {{\sin }^{2}}B-{{\sin }^{2}}C & \cot B-\cot C & 0 \\
   {{\sin }^{2}}C & \cot C & 1 \\
\end{matrix} \right|=\text{0}\]
Now, we know that $\cot x=\dfrac{\cos x}{\sin x}$ , so
\[\left| \begin{matrix}
   {{\sin }^{2}}A-{{\sin }^{2}}B & \dfrac{\operatorname{cosA}}{\operatorname{sinA}}-\dfrac{\operatorname{cosB}}{\operatorname{sinB}} & 0 \\
   {{\sin }^{2}}B-{{\sin }^{2}}C & \dfrac{\operatorname{cosB}}{\operatorname{sinB}}-\dfrac{\operatorname{cosC}}{\operatorname{sinC}} & 0 \\
   {{\sin }^{2}}C & \cot C & 1 \\
\end{matrix} \right|=\text{0}\]
Also, we know that ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$ , so

\[\left| \begin{matrix}
   (\sin A-\sin B)(\sin A+\sin B) & \dfrac{\operatorname{cosA}}{\operatorname{sinA}}-\dfrac{\operatorname{cosB}}{\operatorname{sinB}} & 0 \\
   (\operatorname{sinB}-\operatorname{sinC})(\operatorname{sinB}+\operatorname{sinC}) & \dfrac{\operatorname{cosB}}{\operatorname{sinB}}-\dfrac{\operatorname{cosC}}{\operatorname{sinC}} & 0 \\
   {{\sin }^{2}}C & \cot C & 1 \\
\end{matrix} \right|=\text{0}\]
On simplifying, we get
\[\left| \begin{matrix}
   (\sin A-\sin B)(\sin A+\sin B) & \dfrac{\operatorname{cosA}\operatorname{sinB}-\operatorname{cosB}\operatorname{sinA}}{\operatorname{sinAsinB}} & 0 \\
   (\operatorname{sinB}-\operatorname{sinC})(\operatorname{sinB}+\operatorname{sinC}) & \dfrac{\operatorname{cosB}\operatorname{sinC}-\operatorname{cosCsinB}}{\operatorname{sinB}\operatorname{sinC}} & 0 \\
   {{\sin }^{2}}C & \dfrac{\cos C}{\sin C} & 1 \\
\end{matrix} \right|=\text{0}\]
We know that, $\sin A+\operatorname{sinB}=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
And, $\sin A-\operatorname{sinB}=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$
So, \[\left| \begin{matrix}
   2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)\cdot 2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) & \dfrac{\operatorname{cosA}\operatorname{sinB}-\operatorname{cosB}\operatorname{sinA}}{\operatorname{sinAsinB}} & 0 \\
   2\sin \left( \dfrac{B-C}{2} \right)\cos \left( \dfrac{B+C}{2} \right)\cdot 2\sin \left( \dfrac{B+C}{2} \right)\cos \left( \dfrac{B-C}{2} \right) & \dfrac{\operatorname{cosB}\operatorname{sinC}-\operatorname{cosCsinB}}{\operatorname{sinB}\operatorname{sinC}} & 0 \\
   {{\sin }^{2}}C & \dfrac{\cos C}{\sin C} & 1 \\
\end{matrix} \right|=\text{0}\]
Now, we know that $\sin (A-B)=\sin A\cos B-\cos A\sin B$ , so
\[\left| \begin{matrix}
   2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)\cdot 2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) & \dfrac{\sin (A-B)}{\operatorname{sinAsinB}} & 0 \\
   2\sin \left( \dfrac{B-C}{2} \right)\cos \left( \dfrac{B+C}{2} \right)\cdot 2\sin \left( \dfrac{B+C}{2} \right)\cos \left( \dfrac{B-C}{2} \right) & \dfrac{\sin (B-C)}{\operatorname{sinB}\operatorname{sinC}} & 0 \\
   {{\sin }^{2}}C & \dfrac{\cos C}{\sin C} & 1 \\
\end{matrix} \right|=\text{0}\]
On simplifying, we get
\[\left| \begin{matrix}
   \sin (A+B)\cdot \sin (A-B) & \dfrac{\sin (A-B)}{\operatorname{sinAsinB}} & 0 \\
   \sin (B+C)\cdot \sin (B-C) & \dfrac{\sin (B-C)}{\operatorname{sinB}\operatorname{sinC}} & 0 \\
   {{\sin }^{2}}C & \dfrac{\cos C}{\sin C} & 1 \\
\end{matrix} \right|=\text{0}\]
As, $A+B+C=\pi $
Or, $A+B=\pi -C$
Or, $B+C=\pi -A$
We get,
\[\left| \begin{matrix}
   \operatorname{sinC}\cdot \sin (A-B) & \dfrac{\sin (A-B)}{\operatorname{sinAsinB}} & 0 \\
   \operatorname{sinA}\cdot \sin (B-C) & \dfrac{\sin (B-C)}{\operatorname{sinB}\operatorname{sinC}} & 0 \\
   {{\sin }^{2}}C & \dfrac{\cos C}{\sin C} & 1 \\
\end{matrix} \right|=\text{0}\]
Taking, \[\sin (A-B)\]and \[\sin (B-C)\]common from ${{R}_{1}}$and ${{R}_{2}}$, we get
\[\sin (A-B)\cdot \sin (B-C)\left| \begin{matrix}
   \operatorname{sinC} & \dfrac{1}{\operatorname{sinAsinB}} & 0 \\
   \operatorname{sinA} & \dfrac{1}{\operatorname{sinB}\operatorname{sinC}} & 0 \\
   {{\sin }^{2}}C & \dfrac{\cos C}{\sin C} & 1 \\
\end{matrix} \right|=\text{0}\]
On expanding along column ${{C}_{3}}$,
\[\sin (A-B)\cdot \sin (B-C)\left\{ 0\left( \dfrac{\cos C\operatorname{sinA}}{\sin C}-\dfrac{{{\sin }^{2}}C}{\operatorname{sinB}\operatorname{sinC}} \right)-0\left( \dfrac{\cos C\operatorname{sinC}}{\sin C}-\dfrac{{{\sin }^{2}}C}{\operatorname{sinA}\operatorname{sinB}} \right)+1\left( \dfrac{\sin C}{\operatorname{sinC}\operatorname{sinB}}-\dfrac{\operatorname{sinA}}{\operatorname{sinA}\operatorname{sinB}} \right) \right\}\]on simplifying, we get
\[\sin (A-B)\cdot \sin (B-C)\left\{ \left( \dfrac{1}{\operatorname{sinB}}-\dfrac{1}{\operatorname{sinB}} \right) \right\}\]
On solving we get
$\left| \begin{matrix}
   {{\sin }^{2}}A & \cot A & 1 \\
   {{\sin }^{2}}B & \cot B & 1 \\
   {{\sin }^{2}}C & \cot C & 1 \\
\end{matrix} \right|=\text{0}$

Hence, proved.

Note: In such types of questions multiple concepts are used. To simplify the determinant use elementary operations such that at least two entries in a row or column become 0. This makes calculation easier. All the trigonometric formulas need to be remembered. Also, if we want to calculate the determinant of matrix A of order $3\times 3$, then determinant of matrix A of $3\times 3$ is evaluated as,
$\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$. Try to avoid calculation mistakes.