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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).

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The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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