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In a triangle $\mathrm{ABC}, \mathrm{a}-2 \mathrm{~b}+\mathrm{c}=0 .$ The value of $\cot \left(\frac{\mathrm{A}}{2}\right) \cot \left(\frac{\mathrm{C}}{2}\right)$ is
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3
Given, $\mathrm{a}-2 \mathrm{~b}+\mathrm{c}=0 \Rightarrow \mathrm{a}+\mathrm{c}=2 \mathrm{~b} \cot \left(\frac{\mathrm{A}}{2}\right) \cot \left(\frac{\mathrm{C}}{2}\right)$
$=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{a})}{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \times \frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}}=\sqrt{\frac{\mathrm{s}^{2}}{(\mathrm{~s}-\mathrm{b})^{2}}}=\frac{\mathrm{s}}{\mathrm{s}-\mathrm{b}}$
$=\frac{2 \mathrm{~s}}{2 \mathrm{~s}-2 \mathrm{~b}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}-2 \mathrm{~b}}=\frac{2 \mathrm{~b}+\mathrm{b}}{2 \mathrm{~b}-\mathrm{b}}=\frac{3 \mathrm{~b}}{\mathrm{~b}}=3$
$=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{a})}{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \times \frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}}=\sqrt{\frac{\mathrm{s}^{2}}{(\mathrm{~s}-\mathrm{b})^{2}}}=\frac{\mathrm{s}}{\mathrm{s}-\mathrm{b}}$
$=\frac{2 \mathrm{~s}}{2 \mathrm{~s}-2 \mathrm{~b}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}-2 \mathrm{~b}}=\frac{2 \mathrm{~b}+\mathrm{b}}{2 \mathrm{~b}-\mathrm{b}}=\frac{3 \mathrm{~b}}{\mathrm{~b}}=3$
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