Question

If A, B, C are the angles of a triangle, then
\Delta = \left| {\begin{array}{*{20}{c}} {{{\sin }^2}A}&{\cot A}&1 \\\ {{{\sin }^2}B}&{\cot B}&1 \\\ {{{\sin }^2}C}&{\cot C}&1 \end{array}} \right| = .............

Answer 1

We can expand the determinant along the first column. Then we can change all terms in cot to sines and cosines. Then we can simplify and make it in terms of sines using the identity $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . Then we can use the angle sum property and equate the sine of one angle as the sine of the sum of the other two angles. Then by simplification and cancelling the terms, we can obtain the required value of the determinant.

Complete step by step solution:
We need to find the value of the determinant,
\Delta = \left| {\begin{array}{*{20}{c}} {{{\sin }^2}A}&{\cot A}&1 \\\ {{{\sin }^2}B}&{\cot B}&1 \\\ {{{\sin }^2}C}&{\cot C}&1 \end{array}} \right|
We can expand it along the 1st column, we get,
$\Delta = {\sin ^2}A\left( {\cot B - \cot C} \right) - {\sin ^2}B\left( {\cot A - \cot C} \right) + {\sin ^2}C\left( {\cot A - \cot B} \right)$
We can multiply the middle term with -1 inside and outside the bracket. So, we get,
$\Rightarrow \Delta = {\sin ^2}A\left( {\cot B - \cot C} \right) + {\sin ^2}B\left( {\cot C - \cot A} \right) + {\sin ^2}C\left( {\cot A - \cot B} \right)$ … (1)
Now let us consider the term $\left( {\cot B - \cot C} \right)$. We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$ .
$\Rightarrow \cot B - \cot C = \dfrac{{\cos B}}{{\sin B}} - \dfrac{{\cos C}}{{\sin C}}$
Now we can find the LCM of the denominator,
$\Rightarrow \cot B - \cot C = \dfrac{{\sin C\cos B - \cos C\sin B}}{{\sin B\sin C}}$
We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . So, the above expression will become,
$\Rightarrow \cot B - \cot C = \dfrac{{\sin \left( {C - B} \right)}}{{\sin B\sin C}}$ … (2)
Similarly, consider the 2nd term $\left( {\cot C - \cot A} \right)$
We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$ .
$\Rightarrow \cot C - \cot A = \dfrac{{\cos C}}{{\sin C}} - \dfrac{{\cos A}}{{\sin A}}$
Now we can find the LCM of the denominator,
$\Rightarrow \cot C - \cot A = \dfrac{{\sin A\cos C - \cos A\sin C}}{{\sin C\sin A}}$
We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . So, the above expression will become,
$\Rightarrow \cot C - \cot A = \dfrac{{\sin \left( {A - C} \right)}}{{\sin C\sin A}}$ … (3)
Now we can take the third term, $\left( {\cot A - \cot B} \right)$ . We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$ .
$\Rightarrow \cot A - \cot B = \dfrac{{\cos A}}{{\sin A}} - \dfrac{{\cos B}}{{\sin B}}$
Now we can find the LCM of the denominator,
$\Rightarrow \cot A - \cot B = \dfrac{{\sin B\cos A - \cos B\sin A}}{{\sin A\sin B}}$
We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . So, the above expression will become
$\Rightarrow \cot A - \cot B = \dfrac{{\sin \left( {B - A} \right)}}{{\sin A\sin B}}$ … (4)
Now we can substitute (2) , (3), (4), in (1). Then we will get the determinant as
$\Rightarrow \Delta = {\sin ^2}A\dfrac{{\sin \left( {C - B} \right)}}{{\sin B\sin C}} + {\sin ^2}B\dfrac{{\sin \left( {A - C} \right)}}{{\sin C\sin A}} + {\sin ^2}C\dfrac{{\sin \left( {B - A} \right)}}{{\sin A\sin B}}$
Now we can multiply and divide the terms with $\sin A$ , $\sin B$ and $\sin C$ respectively.
$\Rightarrow \Delta = {\sin ^3}A\dfrac{{\sin \left( {C - B} \right)}}{{\sin A\sin B\sin C}} + {\sin ^3}B\dfrac{{\sin \left( {A - C} \right)}}{{\sin B\sin C\sin A}} + {\sin ^3}C\dfrac{{\sin \left( {B - A} \right)}}{{\sin A\sin B\sin C}}$
Now we can take the common denominator outside.
$\Rightarrow \Delta = \dfrac{1}{{\sin A\sin B\sin C}}\left( {{{\sin }^3}A\sin \left( {C - B} \right) + {{\sin }^3}B\sin \left( {A - C} \right) + {{\sin }^3}C\sin \left( {B - A} \right)} \right)$
Consider the term ${\sin ^3}A\sin \left( {C - B} \right)$ .
It is given that A, B and C are angles of a triangle. So, by angle sum property,
$\Rightarrow A + B + C = 180^\circ$
$\Rightarrow A = 180^\circ - \left( {B + C} \right)$
Then $\sin A = \sin \left( {180^\circ - \left( {B + C} \right)} \right)$
We know that $\sin \left( {180^\circ - x} \right) = \sin x$
$\Rightarrow \sin A = \sin \left( {B + C} \right)$
We can substitute this in the 1st term for one power of $\sin A$
$\Rightarrow {\sin ^3}A\sin \left( {C - B} \right) = {\sin ^2}A\left( {\sin A\sin \left( {C - B} \right)} \right)$
$\Rightarrow {\sin ^3}A\sin \left( {C - B} \right) = {\sin ^2}A\left( {\sin \left( {C + B} \right)\sin \left( {C - B} \right)} \right)$
We know that $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ .
$\Rightarrow {\sin ^3}A\sin \left( {C - B} \right) = {\sin ^2}A\left( {\sin C\cos B + \cos C\sin B} \right)\left( {\sin C\cos B - \cos C\sin B} \right)$
Now this is of the form $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
$\Rightarrow {\sin ^3}A\sin \left( {C - B} \right) = {\sin ^2}A\left[ {{{\left( {\sin C\cos B} \right)}^2} - {{\left( {\cos C\sin B} \right)}^2}} \right]$
Similarly, we can obtain,
$\Rightarrow {\sin ^3}B\sin \left( {A - C} \right) = {\sin ^2}B\left[ {{{\left( {\sin A\cos C} \right)}^2} - {{\left( {\cos A\sin C} \right)}^2}} \right]$ and
$\Rightarrow {\sin ^3}C\sin \left( {B - A} \right) = {\sin ^2}C\left[ {{{\left( {\sin B\cos A} \right)}^2} - {{\left( {\cos B\sin A} \right)}^2}} \right]$
On substituting these relations in the determinant, we get,
$\Rightarrow \Delta = \dfrac{1}{{\sin A\sin B\sin C}}\left( {\sin ^2}A\left[ {{{\left( {\sin C\cos B} \right)}^2} - {{\left( {\cos C\sin B} \right)}^2}} \right] + {\sin ^2}B\left[ {{{\left( {\sin A\cos C} \right)}^2} - {{\left( {\cos A\sin C} \right)}^2}} \right] \\\ \+ {\sin ^2}C\left[ {{{\left( {\sin B\cos A} \right)}^2} - {{\left( {\cos B\sin A} \right)}^2}} \right] \right)$
On expanding the brackets, we get,
$\Rightarrow \Delta = \dfrac{1}{{\sin A\sin B\sin C}}\left( {\sin ^2}A{\sin ^2}C{\cos ^2}B - {\sin ^2}A{\cos ^2}C{\sin ^2}B + {\sin ^2}B{\sin ^2}A{\cos ^2}C - {\sin ^2}B{\cos ^2}A{\sin ^2}C \\\ \+ {\sin ^2}C{\sin ^2}B{\cos ^2}A - {\sin ^2}C{\sin ^2}A{\cos ^2}B \right)$
On further simplification, all the terms inside the brackets cancels out giving the numerator 0.
$\Rightarrow \Delta = \dfrac{1}{{\sin A\sin B\sin C}}\left( 0 \right)$
$\Rightarrow \Delta = 0$

Therefore, the value of the determinant is equal to zero.

Note:
We used the following identities of trigonometry for simplification.
$\cot x = \dfrac{{\cos x}}{{\sin x}}$
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$\sin \left( {180^\circ - x} \right) = \sin x$
As the ratios inside each term vary only in the angles, we only need to do the simplification for one of the terms. We can write the other terms by substituting or changing the angles.