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In a triangle $\mathrm{ABC}$, $\mathrm{if} \cos \mathrm{A}=\cos \mathrm{B} \cos \mathrm{C}$, what is the value of
$\tan A-\tan B-\tan C$ ?
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$\tan A-\tan B-\tan C$ ?
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The correct answer is:
0
As given, $\cos \mathrm{A}=\cos \mathrm{B} \cos \mathrm{C}$
$\tan A-\tan B-\tan C$
$=\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}-\frac{\sin C}{\cos C}$
$=\frac{\sin A}{\cos A}-\frac{(\sin B \cos C+\cos B \sin C)}{\cos B \cos C}$
$=\frac{\sin \mathrm{A}-\sin (\mathrm{B}+\mathrm{C})}{\cos \mathrm{A}} \quad \ldots[\mathrm{using}(1)]$
$\begin{array}{rr}=\frac{\sin A-\sin (\pi-A)}{\cos A} & {[\text { Since, } A+B+C=\pi} \\ & \text { So, } B+C=\pi-A]\end{array}$
$=\frac{\sin A-\sin A}{\cos A}=0$
$\tan A-\tan B-\tan C$
$=\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}-\frac{\sin C}{\cos C}$
$=\frac{\sin A}{\cos A}-\frac{(\sin B \cos C+\cos B \sin C)}{\cos B \cos C}$
$=\frac{\sin \mathrm{A}-\sin (\mathrm{B}+\mathrm{C})}{\cos \mathrm{A}} \quad \ldots[\mathrm{using}(1)]$
$\begin{array}{rr}=\frac{\sin A-\sin (\pi-A)}{\cos A} & {[\text { Since, } A+B+C=\pi} \\ & \text { So, } B+C=\pi-A]\end{array}$
$=\frac{\sin A-\sin A}{\cos A}=0$
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