Question : In the given figure, $ABC$ is an isosceles triangle with $\mathrm{BC}=8 \mathrm{~cm}$ and $\mathrm{AB}=\mathrm{AC}$ $=5 \mathrm{~cm}$. The value of $\tan \mathrm{C}-\cot \mathrm{B}$ is______.
Option 1: $-\frac{5}{12}$
Option 2: $-\frac{7}{12}$
Option 3: $\frac{7}{12}$
Option 4: $\frac{5}{12}$
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Correct Answer: $-\frac{7}{12}$
Solution : Given, $ABC$ is an isosceles triangle with $\mathrm{BC}=8 \mathrm{~cm}$ and $\mathrm{AB}=\mathrm{AC}$ $=5 \mathrm{~cm}$ In $\Delta ABC$, $AD$ is the perpendicular bisector. $BD=DC=\frac{BC}2$ $⇒BD=DC = \frac{8}2 = 4$ In $\Delta ADC$, by Pythagoras Theorem, $AC^2 = DC^2 + AD^2$ $⇒5^2 = 4^2 + AD^2$ $⇒AD^2 = 9$ $\therefore AD = 3$ cm Now, in $\Delta ADC$, $\tan C = \frac{AD}{DC}$ $⇒\tan C = \frac{3}{4}$ Now, in $\Delta ADB$, $\cot B = \frac{BD}{AD}$ $⇒\cot B = \frac{4}{3}$ So, $\tan C - \cot B = \frac{3}{4} - \frac{4}{3}= \frac{9-16}{12}= -\frac{7}{12}$ Hence, the correct answer is $-\frac{7}{12}$.
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