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If \(A, B, C\) are the angles of a triangle such that angle \(A\) is obtuse, then \(\tan B \tan C\) will be less than
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Since, A is obtuse angle \(\therefore 90^{\circ} < \mathrm{A} < 180^{\circ}\)
\(\begin{aligned}
& \Rightarrow 90^{\circ} < 180^{\circ}-(B+C) < 180^{\circ} \\
& \Rightarrow 90^{\circ} > B+C > 0 \Rightarrow B+C < 90^{\circ} \\
& \Rightarrow B < 90^{\circ}-C \Rightarrow \tan B < \tan \left(90^{\circ}-C\right) \\
& \Rightarrow \tan B < \cot C \Rightarrow \tan B \tan C < 1
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow 90^{\circ} < 180^{\circ}-(B+C) < 180^{\circ} \\
& \Rightarrow 90^{\circ} > B+C > 0 \Rightarrow B+C < 90^{\circ} \\
& \Rightarrow B < 90^{\circ}-C \Rightarrow \tan B < \tan \left(90^{\circ}-C\right) \\
& \Rightarrow \tan B < \cot C \Rightarrow \tan B \tan C < 1
\end{aligned}\)
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