Solved Examples - Trigonometric Functions | Mock Tests for JEE Main and Advanced 2025 PDF Download

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Illustration 1: If A + B = π /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 = π /2 and B + C = A.
Hence, A = π /2 – B
So, tan A = tan (π /2 – B) which gives tan A = cot B
Now, tan Atan B = 1
Also, B + C = A
So, C = A – B
tan C = tan (A – B)
also, tan C = (tan A – tanB)/ (1 + tan A tan B)
tan C = (tan A –  tanB)/ (1 + 1)
Therefore, 2 tan C = tan A – 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. (π/3, π/6]
2.(π/2, 2π/3)
3. (2π/3, 5π/6]
4. (5π/3, π]
Solution: We are given that sin A = 1/2 and cos A = 1/3.
Hence, A = π/ 6 and 0 < (cos B = 1/3) <1/2
So, A = π/ 6 and cos-1(0) > B >cos-1(1/2)
Hence, A = π/ 6and π/ 3 <B <π/ 2
π/ 2 <A + B <2π/ 3
Hence, A∈(π/2, 2π/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 = π
Hence, (A + B)/2 = (π– C)/2
So, cot (A + B)/2 = cot (π – C)/2
Hence, {cot A/2. cot B/2 – 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.