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# Notes | EduRev

## JEE Main Mock Test Series 2020 & Previous Year Papers

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## JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course JEE Main Mock Test Series 2020 & Previous Year Papers.
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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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## JEE Main Mock Test Series 2020 & Previous Year Papers

3 videos|174 docs|149 tests

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