If $A+B=45^{\circ}$, then prove that : $(\cot A-1)(\cot B-1)=2$
✓
Answer
$\text {Given,}\quad A+B=45^{\circ} $ $\Rightarrow \quad \cot ( A + B )=\cot 45^{\circ}=1$ Now, $\quad \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}$ $\Rightarrow \quad 1=\frac{\cot A \cot B -1}{\cot A+\cot B }$ $\begin{array}{l}\Rightarrow \quad \cot A +\cot B =\cot A \cot B -1 \\ \Rightarrow \quad \cot A \cot B -\cot A -\cot B =1\end{array}$ Adding 1 on both sides : $\begin{array}{l}\Rightarrow \quad 2=\cot A \cot B-\cot A-\cot B+1 \\\Rightarrow \quad 2=\cot A(\cot B-1)-1(\cot B-1) \\\Rightarrow \quad 2=(\cot A-1)(\cot B-1) \\\Rightarrow \quad(\cot A-1)(\cot B-1)=2\end{array}$
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