Question

If $A+B={225}^o$, then the value of $\left(\frac{\cot A}{1+\cot A}\right).\left(\frac{\cot B}{1+\cot B}\right)$ is

• A. $2$
• B. $1/2$
• C. $3$
• D. $1/3$

Answer 1

The correct option is B $1/2$

$A+B={225}^{\circ }$

$A={225}^{\circ }-B$

$\cot A=\cot ({225}^{\circ }-B)$

$=\cot ({180}^{\circ }+({45}^{\circ }-B))$

$=\cot ({45}^{\circ }-B)$

$=\frac{\cot {45}^{\circ }\cot B+1}{\cot B-\cot {45}^{\circ }}$

$\therefore \cot A=\frac{\cot B+1}{\cot B-1}$

We have $\left(\frac{\cot A}{1+\cot A}\right).\left(\frac{\cot B}{1+\cot B}\right)$

$=\left(\frac{\frac{\cot B+1}{\cot B-1}}{1+\frac{\cot B+1}{\cot B-1}}\right).\left(\frac{\cot B}{1+\cot B}\right)$

$=\left(\frac{\cot B+1}{\cot B-1+\cot B+1}\right).\left(\frac{\cot B}{1+\cot B}\right)$

$=\frac{1}{2}$ on simplification.