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# Solved Examples on Trigonometric Functions

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.