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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