PROBLEMS INVOLVING CONDITIONAL IDENTITIES IN TRIGONOMETRY

Trigonometric identities are true for all admissible values of the angle involved. There are some trigonometric identities which satisfy the given additional conditions. Such identities are called conditional trigonometric identities.

Problem 1 :

If A + B + C = 2s, then prove that

sin(s − A) sin(s − B) + sins sin(s − C)  =  sinA sinB

Solution :

Given : A + B + C  =  2s.

sin(s − A) sin(s − B) :

=  (2/2)sin(s − A) sin(s − B)

=  (1/2)[cos(s - A - s + B) - cos(s - A + s - B)]

=  (1/2)[cos(B - A) - cos(2s - A - B)]

=  (1/2)[cos(B - A) - cos(A + B + C - A - B)]

=  (1/2)[cos(B - A) - cosC]  

sins sin(s − C) :

=  (2/2)sins sin(s − C)

=  (1/2)[2 sins sin(s - C)]

=  (1/2)[cosC - cos(2s - C)]  

sin(s − A) sin(s − B) + sins sin(s − C) :

=  (1/2)[cos(B - A) - cosC] + (1/2)[cosC - cos(2s - C)] 

=  (1/2)[cos(B - A) - cosC + cosC - cos(2s - C)] 

=  (1/2)[cos(B - A) - cos(A + B + C - C)] 

=  (1/2)[cos(B - A) - cos(A + B)] 

=  (1/2)[-2sinB sin(-A)]

=  sinA sin B

Hence proved

Problem 2 :

If x + y + z = xyz, then prove that

(2x/1 − x²) + (2y/1 − y²) + (2z/1 − z²)

=  (2x/1 − x²) (2y/1 − y²) (2z/1 − z²)

Solution :

x + y + z  =  xyz

Let x = tanA, y = tanB and z = tanC.

Then, 

x + y + z  =  xyz

tanA + tanB + tanC  =  tanA tanB tanC

tanA + tanB  =  tanA tanB tanC - tanC

tanA + tanB  =  tanC(tanAtanB - 1)

tanA + tanB  =  -tanC(1 - tanAtanB)

(tanA + tanB) / (1 - tanAtanB)  =  - tanC

tan(A + B)  =  tan(-C)

A + B  =  -C

Multiply each side by 2. 

2A + 2B  =  -2C

tan(2A + 2B) =  tan(- 2C)

(tan2A + tan2B)/(1 - tan2Atan2B)  =  -tan2C

(tan2A + tan2B)  =  -tan2C(1 - tan2Atan2B)

tan2A + tan2B + tan2C  =  tan2Atan2Btan2C -----(1)

tan2A  =  2tanA / 1 - tan²A  =  2x/(1 - x²)

tan2B  =  2tanB / 1 - tan²B  =  2y/(1 - y²)

tan2C  =  2tanC / 1 - tan²C  =  2z/(1 - z²)

Substitute these in (1). 

2x/(1 -x²) + 2y/(1 -y²) + 2z/(1 - z²) 

=  2x/(1 -x²)  2y/(1 -y²) 2z/(1 - z²)

Hence proved.

Problem 3 :

If A + B + C =  π then prove that

sin A + sin B + sin C = 4 cos (A/2) cos (B/2) cos (C/2) 

Solution :

L.H.S

sin A + sin B + sin C

= 2 sin (A + B)/2 cos (A - B)/2 + sin C

Since  A + B + C =  π

A + B = π - C -----(1)

(A + B)/2 = (π - C)/2

= π/2 - C/2

= 2 sin (π/2 - C/2) cos (A - B)/2 + sin C

= 2 cos (C/2) cos (A - B)/2 + sin C

= 2 cos (C/2) cos (A - B)/2 + 2 sin (C/2) cos (C/2)

= 2 cos (C/2) [cos (A - B)/2 + sin (C/2)]

= 2 cos (C/2) [cos (A - B)/2 + sin  (π - (A + B))/2]

= 2 cos (C/2) [cos (A - B)/2 + sin (π/2) -  (A + B)/2]

= 2 cos (C/2) [cos (A - B)/2 + cos (A + B)/2]

= 2 cos (C/2) [2cos ((A - B) + (A + B))/4 cos ((A - B) - (A + B))/4]

= 2 cos (C/2) [2cos ((A - B + A + B)/4 cos (A - B - A - B)/4]

= 2 cos (C/2) [2cos (2A/4) cos (- 2B/4)]

= 2 cos (C/2) [2cos (A/2) cos (B/2)]

= 4 cos (A/2) cos (B/2) cos (C/2)

R.H.S

Hence it is proved

In Δ ABC, prove that

(i) tan A + tan B + tan C = tan A . tan B . tan C

(ii) cot A . cot B + cot B . cot C + cot C . cot A = 1

Solution :

In a triangle, sum of interior angles = 180

A + B + C = π

A + B  = π - C

tan (A + B) = tan (π - C)

(i) tan A + tan B + tan C = tan A . tan B . tan C

(tan A + tan B) / 1 - tan A tan B = tan (π - C)

(tan A + tan B) / 1 - tan A tan B = -tan C

tan A + tan B =  -tan C(1 - tan A tan B)

tan A + tan B =  -tan C + tan A tan B tan C

tan A + tan B + tan C = tan A tan B tan C

ii)  From tan A + tan B + tan C = tan A tan B tan C

(tan A + tan B + tan C)/tan A tan B tan C = 1

(tan A / tan A tan B tan C) + (tan B / tan A tan B tan C ) + (tan C / tan A tan B tan C) = 1

(1/tan B tan C) + (1/tan A tan C ) + (1/tan A tan B) = 1

cot B cot C + cot A cot C + cot A cot B = 1

Hence it is proved

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