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In $\Delta A B C$, if $a^{2} \cos ^{2} A-b^{2}-c^{2}=0,$ then
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The correct answer is:
$\frac{\pi}{2} < A < \pi$
We have, $a^{2} \cos ^{2} A-b^{2}-c^{2}=0$
$\Rightarrow \quad a^{2} \cos ^{2} A=b^{2}+c^{2}$
Also in $\triangle A B C$, we have $\begin{aligned} \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c} &=\frac{a^{2} \cos ^{2} A-a^{2}}{2 b c} \\=& \frac{-a^{2}\left(1-\cos ^{2} A\right)}{2 b c}=\frac{-a^{2} \sin ^{2} A}{2 b c} < 0 \\ & \quad \because 0 < A < \pi, a, b, c>0 \end{aligned}$
$\Rightarrow \cos A < 0 \Rightarrow A$ lies in IInd quadrant.
$\Rightarrow \quad \frac{\pi}{2} < A < \pi$
$\Rightarrow \quad a^{2} \cos ^{2} A=b^{2}+c^{2}$
Also in $\triangle A B C$, we have $\begin{aligned} \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c} &=\frac{a^{2} \cos ^{2} A-a^{2}}{2 b c} \\=& \frac{-a^{2}\left(1-\cos ^{2} A\right)}{2 b c}=\frac{-a^{2} \sin ^{2} A}{2 b c} < 0 \\ & \quad \because 0 < A < \pi, a, b, c>0 \end{aligned}$
$\Rightarrow \cos A < 0 \Rightarrow A$ lies in IInd quadrant.
$\Rightarrow \quad \frac{\pi}{2} < A < \pi$
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