Question

If A + B + C =180° then prove the given conditional trigonometric identity​

Answer 1

Answer:

If A + B + C = π, then the sum of any two angles is supplementary to the third i.e.,

(i) B + C = π - A or, C + A = π - B or A + B = π - C.

(ii) If A + B + C = π then sin (A + B) = sin (π - C) = sin C

sin (B + C) = sin (π - A) = sin A

sin (C + A) = sin (π - B) = sin B

(iii) If A + B + C = π then cos (A + B) = cos (π - C) = - cos C

cos (B + C) = cos (π - A) = - cos A

cos (C + A) = cos (π - B) = - cos B

(iv) If A + B + C = π then tan (A + B) = tan (π - C) = - tan C

tan (B + C) = tan (π - A) = - tan A

tan (C + A) = tan (π - B) = - tan B

(v) If A + B + C = π then A2 + B2 + C2 = π2

Hence, it is evident that the sum of any two of the three angles C2, B2, C2 is complementary to the third.

i.e., A+B2 = π2 - C2,

B+C2 = π2 - A2

C+A2 = π2 - B2

Therefore,

sin (A2 + B2) = sin π2 - C2 = cos C2

sin (B2 + C2) = sin π2 - A2 = cos A2

sin (C2 + A2) = sin π2 - B2 = cos B2

cos (A2 + B2) = cos π2 - C2 = sin C2

sin (B2 + C2) = cos π2 - A2 = sin A2

sin (C2 + A2) = cos π2 - B2 = sin B2

tan (A2 + B2) = tan π2 - C2 = cot C2

tan (B2 + C2) = tan π2 - A2 = cot A2

tan (C2 + A2) = tan π2 - B2 = cot