Match the following
List - I List - II
$1) If A + B = \frac{π}{2} then a) 2 tanAtanB tanB + 2 \tan (A - B) = 2) If A + B = \frac{π}{2} then b) \tan A (1 + tanA) (1 + tanB) = 3) If A + B = \frac{5 π}{4} then c) 2 \frac{\cot}{(1 + \cot A) (1 + \cot B)} = 4) If A + B = \frac{3 π}{4} then (1 + \tan A) (1 + \tan B) = d) \frac{1}{2}$

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d, a, b, c

a, b, c., d

b, c, d, a

c, d, a, b

answer is B.

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Detailed Solution

$\begin{array}{rcl} A + B & = & \frac{π}{2} \\ \tan A \tan B & = & 1 \\ \tan (A - B) & = & \frac{\tan A - \tan B}{1 + \tan A \tan B} \\ \tan (A - B) & = & \frac{\tan A - \tan B}{1 + 1} \\ 2 \tan (A - B) & = & \tan A - \tan B \\ tab B + 2 \tan (A - B) & = & \tan A \end{array}$

$\begin{array}{rcl} A + B & = & \frac{π}{4} \\ (1 + \tan A) (1 + \tan B) & = & 2 \end{array}$

$\begin{array}{rcl} A + B & = & \frac{5 π}{4} \\ \frac{\cot A}{1 + \cot A} \frac{\cot B}{1 + \cot B} & = & \frac{1}{2} \\ A + B & = & \frac{3 π}{4} \\ \tan (A + B) & = & \tan \frac{3 π}{4} \\ \frac{\tan A + \tan B}{1 - \tan A \tan B} & = & - 1 \\ \tan A + \tan B & = & 1 + \tan A \tan B \\ 1 + \tan A + \tan B + \tan A \tan B & = & 2 \tan A \tan B \\ (1 + \tan A) (1 + \tan B) & = & 2 \tan A \tan B \end{array}$