Prove sinA-Bsin A sin B-sinB-Csin B sin C- sinC - Asin C sin A- 0

sin (A B)sin Asin B+sin (B C)sin Bsin C+sin (C A)sin Csin A=0 =sin Acos B cos Asin Bsin Asin B+sin Bcos C cos Bsin Csin Bsin C+sin Ccos A cos Csin Asi ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

Prove sinA-...

Question

Prove: $\frac{sin (A - B)}{sin A sin B} + \frac{sin (B - C)}{sin B sin C} + \frac{sin (C - A)}{sin C sin A} = 0$

Open in App

Solution

Given: $\frac{sin (A - B)}{sin A sin B} + \frac{sin (B - C)}{sin B sin C} + \frac{sin (C - A)}{sin C sin A} = 0$
LHS: $\frac{sin (A - B)}{sin A sin B} + \frac{sin (B - C)}{sin B sin C} + \frac{sin (C - A)}{sin C sin A}$
We know that $sin (α + β) = sin α cos β - cos α sin β$
$= \frac{sin A cos B - cos A sin B}{sin A sin B} + \frac{sin B cos C - cos B sin C}{sin B sin C} + \frac{sin C cos A - cos C sin A}{sin C sin A}$
$= \frac{sin A cos B}{sin A sin B} - \frac{cos A sin B}{sin A sin B} + \frac{sin B cos C}{sin B sin C} - \frac{cos B sin C}{sin B sin C} + \frac{sin C cos A}{sin C sin A} - \frac{cos C sin A}{sin C sin A}$
$= cot B - cot A + cos C - cot B + cot A - cot C$
$= 0$
$=$ RHS
Hence proved.

Suggest Corrections

Similar questions

Q. Show that:
(i) sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0
(ii) sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0

Q. If

A + B + C = π

, prove that

(sin A + sin B + sin C) (- sin A + sin B + sin C) (sin A - sin B + sin C) (sin A + sin B - sin C) = 4 {sin}^{2} A {sin}^{2} B {sin}^{2} C

Q. Prove that:
(i)

\frac{s i n (A + B) + s i n (A - B)}{c o s (A + B) + c o s (A - B)} = t a n A

(ii)

\frac{s i n (A - B)}{c o s A c o s B} + \frac{s i n (B - C)}{c o s B c o s C} + \frac{s i n (C - A)}{c o s C c o s A} = 0

(iii)

\frac{s i n (A - B)}{s i n A s i n B} + \frac{s i n (B - C)}{s i n B s i n C} + \frac{s i n (C - A)}{s i n C s i n A} = 0

(iv)

s i n^{2} B = s i n^{2} A + s i n^{2} (A - B) - 2 s i n A c o s B s i n (A - B)

(v)

c o s^{2} + c o s^{2} B - 2 c o s A c o s B c o s (A + B) = s i n^{2} (A + B)

(vi)

\frac{t a n (A + B)}{c o t (A - B)} = \frac{t a n^{2} A - t a n^{2} B}{1 - t a n^{2} A t a n^{2} B}

Q. Prove that

\frac{a^{2} sin (B - C)}{sin B + sin C} + \frac{b^{2} sin (C - A)}{sin C + sin A} + \frac{c^{2} sin (A - B)}{sin A + sin B} = 0

\frac{cos A}{sin B sin C} + \frac{cos B}{sin C sin A} + \frac{cos C}{sin A sin B} = 2

Join BYJU'S Learning Program

Prove: sin(A−B)sinAsinB+sin(B−C)sinBsinC+sin(C−A)sinCsinA=0

Prove: $\frac{sin (A - B)}{sin A sin B} + \frac{sin (B - C)}{sin B sin C} + \frac{sin (C - A)}{sin C sin A} = 0$