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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

In $\Delta ABC$,

$\left| {\begin{array}{*{20}{c}}{\cos 3A}&{\cos 3B}&{\cos 3C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$

Select the right option

(A) Triangle is isosceles
(B) Triangle is equilateral
(C) Triangle is right angled
(D) Triangle is scalene

We have, $\left| {\begin{array}{*{20}{c}}{4{{\cos }^3}A - 3\cos A}&{4{{\cos }^3}B - 3\cos B}&{4{{\cos }^3}C - 3\cos C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}{4{{\cos }^3}A}&{4{{\cos }^3}B}&{4{{\cos }^3}C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| - \left| {\begin{array}{*{20}{c}}{3\cos A}&{3\cos B}&{3\cos C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}{{{\cos }^3}A}&{{{\cos }^3}B}&{{{\cos }^3}C}\\{\cos A}&{\cos B}&{\cos C}\\{\sin A}&{\sin B}&{\sin C}\end{array}} \right| = 0$ as the 2nd determinant is 0.

$ \Rightarrow \cos A.\cos B.\cos C\left| {\begin{array}{*{20}{c}}{{{\cos }^2}A}&{{{\cos }^2}B}&{{{\cos }^2}C}\\1&1&1\\{\tan A}&{\tan B}&{\tan C}\end{array}} \right| = 0$

$ \Rightarrow \cos A\cos B\cos C{\left| {\begin{array}{*{20}{c}}{{{\cos }^2}A - {{\cos }^2}C}&{{{\cos }^2}B - {{\cos }^2}C}&{{{\cos }^2}C}\\0&0&1\\{\tan A - \tan C}&{\tan B - \tan C}&{\tan C}\end{array}} \right|_{\scriptstyle{C_1} \to {C_1} - {C_3}\atop\scriptstyle{C_2} \to {C_2} - {C_3}}} = 0$

Since 0 is on the right side, while the constants & negative signs etc. can be removed from the left side, cosA.cosB.cosC should not be removed.

$ \Rightarrow \cos A\cos B\cos C\left| {\begin{array}{*{20}{c}}{{{\cos }^2}A - {{\cos }^2}C}&{{{\cos }^2}B - {{\cos }^2}C}\\{\tan A - \tan C}&{\tan B - \tan C}\end{array}} \right| = 0$

$ \Rightarrow \cos A\cos B\cos C[({\cos ^2}A - {\cos ^2}C)(\tan B - \tan C) - ({\cos ^2}B - {\cos ^2}C)(\tan A - \tan C)] = 0$

$ \Rightarrow \cos A\sin (C + A)\sin (C - A)\sin (B - C) - \cos B\sin (C + B)\sin (C - B)\sin (A - C) = 0$

$ \Rightarrow \cos A\sin B\sin (C - A)\sin (B - C) - \cos B\sin A\sin (B - C)\sin (C - A) = 0$

$ \Rightarrow \sin (B - C)\sin (C - A)\left( {\sin B\cos A - \sin A\cos B} \right) = 0$

$ \Rightarrow \sin (A - B)\sin (B - C)\sin (C - A) = 0$

Clearly, a pair of angles must be the same.

Option (A).