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**Illustration 1: If A + B = Ï€ /2 and B + C = A, then tan A equals (2001)****1. 2 (tan B + tan C)****2. tan B + tan C****3. tan B + 2tan C****4. 2 tan B + tan C****Solution: **It is given in the question that A + B = Ï€ /2 and B + C = A.

Hence, A = Ï€ /2 â€“ B

So, tan A = tan (Ï€ /2 â€“ B) which gives tan A = cot B

Now, tan Atan B = 1

Also, B + C = A

So, C = A â€“ B

tan C = tan (A â€“ B)

also, tan C = (tan A â€“ tanB)/ (1 + tan A tan B)

tan C = (tan A â€“ tanB)/ (1 + 1)

Therefore, 2 tan C = tan A â€“ tan B

Hence, tan A =tanB + 2 tan C

**Illustration 2: Given both A and B are acute angles, sin A = 1/2 and cos A = 1/3, then the value of A+B belongs to (2004)****1. (Ï€/3, Ï€/6]****2.(Ï€/2, 2Ï€/3)****3. (2Ï€/3, 5Ï€/6]****4. (5Ï€/3, Ï€]****Solution: **We are given that sin A = 1/2 and cos A = 1/3.

Hence, A = Ï€/ 6 and 0 < (cos B = 1/3) <1/2

So, A = Ï€/ 6 and cos-1(0) > B >cos-1(1/2)

Hence, A = Ï€/ 6and Ï€/ 3 <B <Ï€/ 2

Ï€/ 2 <A + B <2Ï€/ 3

Hence, Aâˆˆ(Ï€/2, 2Ï€/3).

**Illustration 3: In any triangle, prove that****cot A/2 + cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2. (2000)****Solution:**We know that A + B + C = Ï€

Hence, (A + B)/2 = (Ï€â€“ C)/2

So, cot (A + B)/2 = cot (Ï€ â€“ C)/2

Hence, {cot A/2. cot B/2 â€“ 1}/ {cot A/2 + cot B/2} = tan C/2

Hence, cot A/2 + cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2.

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