Solveeit Logo
Login

Question

Question: If \(A+B+C=\pi \) , prove that \(\left| \begin{matrix} {{\sin }^{2}}A & \cot A & 1 \\\ {{\...

If A+B+C=πA+B+C=\pi , prove that ∣sin⁡2Acot⁡A1 sin⁡2Bcot⁡B1 sin⁡2Ccot⁡C1 ∣=0\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}

Explanation

Solution

To solve this question we will first using elementary row operation R1→R1−R2{{R}_{1}}\to {{R}_{1}}-{{R}_{2}} and R2→R2−R3{{R}_{2}}\to {{R}_{2}}-{{R}_{3}} then we will us trigonometric identities which are cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}, sin⁡A+sinB⁡=2sin⁡(A+B2)cos⁡(A−B2)\sin A+\operatorname{sinB}=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right), sin⁡A−sinB⁡=2sin⁡(A−B2)cos⁡(A+B2)\sin A-\operatorname{sinB}=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right) and algebraic identity a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=(a-b)(a+b)to solve the determinant. At last we will expand determinant along column C3{{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 3×33\times 3, then determinant of matrix A of 3×33\times 3 is evaluated as,
∣a11a12a13 a21a22a23 a31a32a33 ∣=a11(a22a33−a32a23)−a21(a12a33−a32a13)+a31(a23a12−a22a13)\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=πA+B+C=\pi and asked to prove that ∣sin⁡2Acot⁡A1 sin⁡2Bcot⁡B1 sin⁡2Ccot⁡C1 ∣=0\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 R1→R1−R2{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}, we get

{{\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.