Match the following List - I List - II1) If A+B=π2 then a) 2tanAtanB tanB+2tan(A-B)= 2) If A+B=π2 then b) tan A (1+tanA) (1+tanB)= 3) If A+B=5π4 then c) 2 cot(1+cot A) (1+cot B)= 4) If A+B=3π4 then (1+tan A) (1+tan B)= d) 12

1)A+B=π2tan A tan B=1tan (A-B)=tan A-tan B1+tan A tan Btan (A-B)=tan A-tan B1+12tan(A-B)=tan A-tan Btab B+2tan(A-B)=tan A 2)A+B=π4(1+tan A) (1+tan B)=2 3)A+B=5π4cot A1+cot A cot B1+cot B=12A+B=3π4tan (A+B)=tan 3π4tan A+tan B1-tan A tan B=-1tan A+tan B=1+tan A tan B1+tan A+tan B+tan A tan B=2 tan A tan B(1+tan A) (1+tan B)=2 tan A tan B

Match the following List - I List - II1) If A+B=π2 then a) 2tanAtanB tanB+2tan(A-B)= 2) If A+B=π2 then b) tan A (1+tanA) (1+tanB)= 3) If A+B=5π4 then c) 2 cot(1+cot A) (1+cot B)= 4) If A+B=3π4 then (1+tan A) (1+tan B)= d) 12

1)A+B=π2tan A tan B=1tan (A-B)=tan A-tan B1+tan A tan Btan (A-B)=tan A-tan B1+12tan(A-B)=tan A-tan Btab B+2tan(A-B)=tan A 2)A+B=π4(1+tan A) (1+tan B)=2 3)A+B=5π4cot A1+cot A cot B1+cot B=12A+B=3π4tan (A+B)=tan 3π4tan A+tan B1-tan A tan B=-1tan A+tan B=1+tan A tan B1+tan A+tan B+tan A tan B=2 tan A tan B(1+tan A) (1+tan B)=2 tan A tan B

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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

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

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

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