A) 1
B) \[-1\]
C) 0
D) 2
Correct Answer: D
Solution :
As, \[(\cot A-1)(\cot B-1)\] \[=\left( \frac{1}{\tan \,A}-1 \right)\left( \frac{1}{\tan B}-1 \right)\] \[=\left( \frac{1}{\tan A}-1 \right)\left( \frac{1}{tan(45{}^\circ -A)}-1 \right)\] \[(\because A+B=45{}^\circ )\] \[=\left( \frac{1-\tan A}{\tan A} \right)\left( \frac{1+\tan 45{}^\circ \tan A}{\tan 45{}^\circ -\tan A}-1 \right)\] \[=\left( \frac{1-\tan A}{\tan A} \right)\left( \frac{1+\tan A}{1-\tan A}-1 \right)\] \[=\left( \frac{1-\tan A}{\tan A} \right)\left( \frac{2\tan A}{1-\tan A} \right)=2\]You need to login to perform this action.
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