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### 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).

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$