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
Verified640.8k+ views
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.
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.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE
Draw a labelled diagram of the human male reproductive class 12 biology CBSE
Describe the experiment to compare the emf of two cells class 12 physics CBSE
What is standard hydrogen electrode
What is conventional current and electric current class 12 physics CBSE
2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE
Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
A member of Simon commission later became Prime Minister class 12 social science CBSE
Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE
Give one example of a reptile that is viviparous class 12 biology CBSE
Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE