In any $\Delta ABC$, prove that: $\left( {{b}^{2}}-{{c}^{2}} \right)\cot A+\left( {{c}^{2}}-{{a}^{2}} \right)\cot B+\left( {{a}^{2}}-{{b}^{2}} \right)\cot C=0$
Answer
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Hint: To prove the given equation we will use Sine Rule or Law of Sines. Firstly we will write the Sine rule then using it we will substitute the value in on the left hand side of the equation such that we can cancel out all the terms and get the value equal to the right hand side and hence prove our equation.
Complete step by step answer:
We have to prove that:
$\left( {{b}^{2}}-{{c}^{2}} \right)\cot A+\left( {{c}^{2}}-{{a}^{2}} \right)\cot B+\left( {{a}^{2}}-{{b}^{2}} \right)\cot C=0$….$\left( 1 \right)$
Now we will take left hand side value and simplify it to get right hand side values as follows:
So we have the left hand side value as:
$\left( {{b}^{2}}-{{c}^{2}} \right)\cot A+\left( {{c}^{2}}-{{a}^{2}} \right)\cot B+\left( {{a}^{2}}-{{b}^{2}} \right)\cot C$….$\left( 2 \right)$
We know the Sine rule is given as:
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$
Where $A,B,C$ are the angles of the triangle and $a,b,c$ are the side of the triangle.
So as we know that sum of all the angles of the triangle is $\pi $ which means:
$A+B+C=\pi $……$\left( 3 \right)$
From above we can get value of $a,b,c$ as below:
$\begin{align}
& a=2R\sin A \\
& b=2R\sin B \\
& c=2R\sin C \\
\end{align}$
Substitute above value in equation (2) and simplify as follows:
$\begin{align}
& \left( {{\left( 2R\sin B \right)}^{2}}-{{\left( 2R\sin C \right)}^{2}} \right)\cot A+\left( {{\left( 2R\sin C \right)}^{2}}-{{\left( 2R\sin A \right)}^{2}} \right)\cot B+\left( {{\left( 2R\sin A \right)}^{2}}-{{\left( 2R\sin B \right)}^{2}} \right)\cot C \\
& = \left( 4{{R}^{2}}{{\sin }^{2}}B-4{{R}^{2}}{{\sin }^{2}}C \right)\cot A+\left( 4{{R}^{2}}{{\sin }^{2}}C-4{{R}^{2}}{{\sin }^{2}}A \right)\cot B+\left( 4{{R}^{2}}{{\sin }^{2}}A-4{{R}^{2}}{{\sin }^{2}}B \right)\cot C \\
\end{align}$
Now we will use the below formula in above equation by multiplying and dividing each term by 2:
\[2{{\sin }^{2}}X=1-\cos 2X\]
So we get,
$\begin{align}
& = \left( 4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}B-4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}C \right)\cot A+\left( 4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}C-4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}A \right)\cot B+\left( 4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}A-4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}B \right)\cot C \\
& = \left( 4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2B \right)-4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2C \right) \right)\cot A+\left( 4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2C \right)-4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2A \right) \right)\cot B+\left( 4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2A \right)-4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2B \right) \right)\cot C \\
\end{align}$
Now take $4{{R}^{2}}$common and simplify the brackets as follows:
$\begin{align}
& = 4{{R}^{2}}\left( \left( \dfrac{1}{2}\left( \left( 1-\cos 2B \right)-\left( 1-\cos 2C \right) \right) \right)\cot A+\left( \dfrac{1}{2}\left( \left( 1-\cos 2C \right)-\left( 1-\cos 2A \right) \right) \right)\cot B+\left( \dfrac{1}{2}\left( \left( 1-\cos 2A \right)-\left( 1-\cos 2B \right) \right) \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \left( \dfrac{1}{2}\left( 1-\cos 2B-1+\cos 2C \right) \right)\cot A+\left( \dfrac{1}{2}\left( 1-\cos 2C-1+\cos 2A \right) \right)\cot B+\left( \dfrac{1}{2}\left( 1-\cos 2A-1+\cos 2B \right) \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \left( \dfrac{1}{2}\left( \cos 2C-\cos 2B \right) \right)\cot A+\left( \dfrac{1}{2}\left( \cos 2A-\cos 2C \right) \right)\cot B+\left( \dfrac{1}{2}\left( \cos 2B-\cos 2A \right) \right)\cot C \right) \\
\end{align}$
Now we will use the below formula in above equation:
$\left( \cos A-\cos B \right)=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)$
So we get,
$\begin{align}
& = 4{{R}^{2}}\left( \left( \dfrac{1}{2}\left( \cos 2C-\cos 2B \right) \right)\cot A+\left( \dfrac{1}{2}\left( \cos 2A-\cos 2C \right) \right)\cot B+\left( \dfrac{1}{2}\left( \cos 2B-\cos 2A \right) \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \dfrac{1}{2}\times 2\sin \left( \dfrac{2B+2C}{2} \right)\sin \left( \dfrac{2B-2C}{2} \right)\cot A+\dfrac{1}{2}\times 2\sin \left( \dfrac{2C+2A}{2} \right)\sin \left( \dfrac{2C-2A}{2} \right)\cot B+\dfrac{1}{2}\times 2\sin \left( \dfrac{2A+2B}{2} \right)\sin \left( \dfrac{2A-2B}{2} \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \sin \left( B+C \right)\sin \left( B-C \right)\cot A+\sin \left( C+A \right)\sin \left( C-A \right)\cot B+\sin \left( A+B \right)\sin \left( A-B \right)\cot C \right) \\
\end{align}$
Next we will use the value from equation (3) above and get,
$\begin{align}
& = 4{{R}^{2}}\left( \sin \left( \pi -A \right)\sin \left( B-C \right)\cot A+\sin \left( \pi -B \right)\sin \left( C-A \right)\cot B+\sin \left( \pi -C \right)\sin \left( A-B \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \sin A\sin \left( B-C \right)\cot A+\sin B\sin \left( C-A \right)\cot B+\sin C\sin \left( A-B \right)\cot C \right) \\
\end{align}$
Now we will expand the cotangent term in sine and cosine term by using the below relation:
$\cot X=\dfrac{\cos X}{\sin X}$
So we get,
$\begin{align}
& = 4{{R}^{2}}\left( \sin A\sin \left( B-C \right)\times \dfrac{\cos A}{\sin A}+\sin B\sin \left( C-A \right)\times \dfrac{\cos B}{\sin B}+\sin C\sin \left( A-B \right)\times \dfrac{\cos C}{\sin C} \right) \\
& = 4{{R}^{2}}\left( \sin \left( B-C \right)\cos A+\sin \left( C-A \right)\cos B+\sin \left( A-B \right)\cos C \right) \\
\end{align}$
Now use the below in above value,
$\sin \left( X-Y \right)=\sin X\cos Y-\cos X\sin Y$
So we get,
$\begin{align}
& = 4{{R}^{2}}\left( \left( \sin B\cos C-\cos B\sin C \right)\cos A+\left( \sin C\cos A-\cos C\sin A \right)\cos B+\left( \sin A\cos B-\cos A\sin B \right)\cos C \right) \\
& = 4{{R}^{2}}\left( \sin B\cos C\cos A-\cos B\sin C\cos A+\sin C\cos A\cos B-\cos C\sin A\cos B+\sin A\cos B\cos C-\cos A\sin B\cos C \right) \\
& = 4{{R}^{2}}\times 0 \\
& = 0 \\
\end{align}$
We got the right hand side value.
Hence proved that $\left( {{b}^{2}}-{{c}^{2}} \right)\cot A+\left( {{c}^{2}}-{{a}^{2}} \right)\cot B+\left( {{a}^{2}}-{{b}^{2}} \right)\cot C=0$
Note: Trigonometry is a very important topic of mathematics which deals with the sides and angle of triangles. There are six trigonometric functions namely sine, cosine, tangent, secant, cosecant, cotangent. There are many relations among these functions and there are various rules and properties of them. Sine law or Sine rule is the equation which relates the sides of the triangles to sides of its angles.
Complete step by step answer:
We have to prove that:
$\left( {{b}^{2}}-{{c}^{2}} \right)\cot A+\left( {{c}^{2}}-{{a}^{2}} \right)\cot B+\left( {{a}^{2}}-{{b}^{2}} \right)\cot C=0$….$\left( 1 \right)$
Now we will take left hand side value and simplify it to get right hand side values as follows:
So we have the left hand side value as:
$\left( {{b}^{2}}-{{c}^{2}} \right)\cot A+\left( {{c}^{2}}-{{a}^{2}} \right)\cot B+\left( {{a}^{2}}-{{b}^{2}} \right)\cot C$….$\left( 2 \right)$
We know the Sine rule is given as:
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$
Where $A,B,C$ are the angles of the triangle and $a,b,c$ are the side of the triangle.
So as we know that sum of all the angles of the triangle is $\pi $ which means:
$A+B+C=\pi $……$\left( 3 \right)$
From above we can get value of $a,b,c$ as below:
$\begin{align}
& a=2R\sin A \\
& b=2R\sin B \\
& c=2R\sin C \\
\end{align}$
Substitute above value in equation (2) and simplify as follows:
$\begin{align}
& \left( {{\left( 2R\sin B \right)}^{2}}-{{\left( 2R\sin C \right)}^{2}} \right)\cot A+\left( {{\left( 2R\sin C \right)}^{2}}-{{\left( 2R\sin A \right)}^{2}} \right)\cot B+\left( {{\left( 2R\sin A \right)}^{2}}-{{\left( 2R\sin B \right)}^{2}} \right)\cot C \\
& = \left( 4{{R}^{2}}{{\sin }^{2}}B-4{{R}^{2}}{{\sin }^{2}}C \right)\cot A+\left( 4{{R}^{2}}{{\sin }^{2}}C-4{{R}^{2}}{{\sin }^{2}}A \right)\cot B+\left( 4{{R}^{2}}{{\sin }^{2}}A-4{{R}^{2}}{{\sin }^{2}}B \right)\cot C \\
\end{align}$
Now we will use the below formula in above equation by multiplying and dividing each term by 2:
\[2{{\sin }^{2}}X=1-\cos 2X\]
So we get,
$\begin{align}
& = \left( 4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}B-4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}C \right)\cot A+\left( 4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}C-4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}A \right)\cot B+\left( 4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}A-4{{R}^{2}}\times \dfrac{2}{2}{{\sin }^{2}}B \right)\cot C \\
& = \left( 4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2B \right)-4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2C \right) \right)\cot A+\left( 4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2C \right)-4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2A \right) \right)\cot B+\left( 4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2A \right)-4{{R}^{2}}\times \dfrac{1}{2}\left( 1-\cos 2B \right) \right)\cot C \\
\end{align}$
Now take $4{{R}^{2}}$common and simplify the brackets as follows:
$\begin{align}
& = 4{{R}^{2}}\left( \left( \dfrac{1}{2}\left( \left( 1-\cos 2B \right)-\left( 1-\cos 2C \right) \right) \right)\cot A+\left( \dfrac{1}{2}\left( \left( 1-\cos 2C \right)-\left( 1-\cos 2A \right) \right) \right)\cot B+\left( \dfrac{1}{2}\left( \left( 1-\cos 2A \right)-\left( 1-\cos 2B \right) \right) \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \left( \dfrac{1}{2}\left( 1-\cos 2B-1+\cos 2C \right) \right)\cot A+\left( \dfrac{1}{2}\left( 1-\cos 2C-1+\cos 2A \right) \right)\cot B+\left( \dfrac{1}{2}\left( 1-\cos 2A-1+\cos 2B \right) \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \left( \dfrac{1}{2}\left( \cos 2C-\cos 2B \right) \right)\cot A+\left( \dfrac{1}{2}\left( \cos 2A-\cos 2C \right) \right)\cot B+\left( \dfrac{1}{2}\left( \cos 2B-\cos 2A \right) \right)\cot C \right) \\
\end{align}$
Now we will use the below formula in above equation:
$\left( \cos A-\cos B \right)=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)$
So we get,
$\begin{align}
& = 4{{R}^{2}}\left( \left( \dfrac{1}{2}\left( \cos 2C-\cos 2B \right) \right)\cot A+\left( \dfrac{1}{2}\left( \cos 2A-\cos 2C \right) \right)\cot B+\left( \dfrac{1}{2}\left( \cos 2B-\cos 2A \right) \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \dfrac{1}{2}\times 2\sin \left( \dfrac{2B+2C}{2} \right)\sin \left( \dfrac{2B-2C}{2} \right)\cot A+\dfrac{1}{2}\times 2\sin \left( \dfrac{2C+2A}{2} \right)\sin \left( \dfrac{2C-2A}{2} \right)\cot B+\dfrac{1}{2}\times 2\sin \left( \dfrac{2A+2B}{2} \right)\sin \left( \dfrac{2A-2B}{2} \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \sin \left( B+C \right)\sin \left( B-C \right)\cot A+\sin \left( C+A \right)\sin \left( C-A \right)\cot B+\sin \left( A+B \right)\sin \left( A-B \right)\cot C \right) \\
\end{align}$
Next we will use the value from equation (3) above and get,
$\begin{align}
& = 4{{R}^{2}}\left( \sin \left( \pi -A \right)\sin \left( B-C \right)\cot A+\sin \left( \pi -B \right)\sin \left( C-A \right)\cot B+\sin \left( \pi -C \right)\sin \left( A-B \right)\cot C \right) \\
& = 4{{R}^{2}}\left( \sin A\sin \left( B-C \right)\cot A+\sin B\sin \left( C-A \right)\cot B+\sin C\sin \left( A-B \right)\cot C \right) \\
\end{align}$
Now we will expand the cotangent term in sine and cosine term by using the below relation:
$\cot X=\dfrac{\cos X}{\sin X}$
So we get,
$\begin{align}
& = 4{{R}^{2}}\left( \sin A\sin \left( B-C \right)\times \dfrac{\cos A}{\sin A}+\sin B\sin \left( C-A \right)\times \dfrac{\cos B}{\sin B}+\sin C\sin \left( A-B \right)\times \dfrac{\cos C}{\sin C} \right) \\
& = 4{{R}^{2}}\left( \sin \left( B-C \right)\cos A+\sin \left( C-A \right)\cos B+\sin \left( A-B \right)\cos C \right) \\
\end{align}$
Now use the below in above value,
$\sin \left( X-Y \right)=\sin X\cos Y-\cos X\sin Y$
So we get,
$\begin{align}
& = 4{{R}^{2}}\left( \left( \sin B\cos C-\cos B\sin C \right)\cos A+\left( \sin C\cos A-\cos C\sin A \right)\cos B+\left( \sin A\cos B-\cos A\sin B \right)\cos C \right) \\
& = 4{{R}^{2}}\left( \sin B\cos C\cos A-\cos B\sin C\cos A+\sin C\cos A\cos B-\cos C\sin A\cos B+\sin A\cos B\cos C-\cos A\sin B\cos C \right) \\
& = 4{{R}^{2}}\times 0 \\
& = 0 \\
\end{align}$
We got the right hand side value.
Hence proved that $\left( {{b}^{2}}-{{c}^{2}} \right)\cot A+\left( {{c}^{2}}-{{a}^{2}} \right)\cot B+\left( {{a}^{2}}-{{b}^{2}} \right)\cot C=0$
Note: Trigonometry is a very important topic of mathematics which deals with the sides and angle of triangles. There are six trigonometric functions namely sine, cosine, tangent, secant, cosecant, cotangent. There are many relations among these functions and there are various rules and properties of them. Sine law or Sine rule is the equation which relates the sides of the triangles to sides of its angles.
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