Prove that- sin A-B sin A sin B -sin B-C sin B sin C -sin C-A sin C sin A -0

sin (A B) #160; sin A sin B #160; =sin A cos B #160; sin A sin B #160; sin B cos A #160; sin A sin B #160; = B A LHS= ( B ...

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Standard XII

Mathematics

Integration by Partial Fractions

Prove that: ...

Question

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

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Solution

$\frac{sin (A - B)}{sin A sin B} = \frac{sin A cos B}{sin A sin B} - \frac{sin B cos A}{sin A sin B} = cot B - cot A$
LHS= $(cot B - cot A) + (cot C - cot B) + (cot A - cot C) = 0$

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Similar questions

Show that :
(i) $s i n A s i n (B — C) + s i n B s i n (C — A) + s i n C s i n (A — B) = 0$
(ii) $s i n (B — C) c o s (A — D) + s i n (C — A) c o s (B — D) + s i n (A — B) c o s (C — D) = 0$

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:

sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = 4 sinAsinBsinC

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

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

sin(A−B)sinAsinB=sinAcosBsinAsinB−sinBcosAsinAsinB=cotB−cotALHS=(cotB−cotA)+(cotC−cotB)+(cotA−cotC)=0

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

$\frac{sin (A - B)}{sin A sin B} = \frac{sin A cos B}{sin A sin B} - \frac{sin B cos A}{sin A sin B} = cot B - cot A$
LHS= $(cot B - cot A) + (cot C - cot B) + (cot A - cot C) = 0$