A) 1
B) - 1
C) 0
D) 1/2
Correct Answer: D
Solution :
\[\frac{\cot A}{1+\cot A}\,.\,\frac{\cot B}{1+\cot B}=\frac{1}{(1+\tan A)\,(1+\tan B)}\] \[=\frac{1}{\tan A+\tan B+1+\tan A\tan B}\] \[[\because \,\tan (A+B)=\tan {{225}^{o}}]\] \[\Rightarrow \,\tan \,A+\tan \,B=1-\tan \,A\,\tan B]\] \[=\frac{1}{1-\tan A\,\tan B+1+\tan A\tan B}=\frac{1}{2}\].
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