Question

If A + B = ${225}^{\circ }$, then $\frac{cotA}{1+cotA}$. $\frac{cotB}{1+cotB}$ =

• A. 1
• B. -1
• C. 0
• D. $\frac{1}{2}$

Answer 1

The correct option is D

$\frac{1}{2}$

$\frac{cotA}{1+cotA}$. $\frac{cotB}{1+cotB}$ = $\frac{1}{(1+tanA)(1+tanB)}$

= $\frac{1}{tanA+tanB+1+tanAtanB}$

[ ∵ tan(A + B) = $tan{225}^{\circ }$]

$\Rightarrow$ tanA + tan B = 1 - tan A tan B

= $\frac{1}{1-tanAtanB+1+tanAtanB}$ = $\frac{1}{2}$.