Question

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

• A. 1
• B. – 1
• C. 0
• D. ½

Answer 1

Answer: (D)

½

Answer 2

$\frac{\cot A}{1 + \cot A}.\frac{\cot B}{1 + \cot B} = \frac{1}{(1 + \tan A)(1 + \tan B)}$

$= \frac{1}{\tan A + \tan B + 1 + \tan A\tan B}$ $\lbrack\because\tan(A + B) = \tan 225^{o}\rbrack$

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

$= \frac{1}{1 - \tan A\tan B + 1 + \tan A\tan B} = \frac{1}{2}$.