Advertisements
Advertisements
प्रश्न
Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
Advertisements
उत्तर
Consider LHS:
\[\sin \left( B - C \right) \cos \left( A - D \right) + \sin \left( C - A \right) \cos \left( B - D \right) + \sin \left( A - B \right) \cos \left( C - D \right)\]
\[= \frac{1}{2}\left[ 2\sin \left( B - C \right) \cos \left( A - D \right) \right] + \frac{1}{2}\left[ 2\sin \left( C - A \right) \cos \left( B - D \right) \right] + \frac{1}{2}\left[ 2\sin \left( A - B \right) \cos\left( C - D \right) \right]\]
\[ = \frac{1}{2}\left[ \sin \left\{ \left( B - C \right) + \left( A - D \right) \right\} + \sin \left\{ \left( B - C \right) - \left( A - D \right) \right\} \right] + \frac{1}{2}\left[ \sin \left\{ \left( C - A \right) + \left( B - D \right) \right\} + \sin \left\{ \left( C - A \right) - \left( B - D \right) \right\} \right] + \frac{1}{2}\left[ \sin \left\{ \left( A - B \right) + \left( C - D \right) \right\} + \sin \left\{ \left( A - B \right) - \left( C - D \right) \right\} \right]\]
\[ = \frac{1}{2}\left[ \sin \left( B - C + A - D \right) + \sin \left( B - C - A + D \right) \right] + \frac{1}{2}\left[ \sin \left( C - A + B - D \right) + \sin \left( C - A - B + D \right) \right] + \frac{1}{2}\left[ \sin \left( A - B + C - D \right) + \sin \left( A - B - C + D \right) \right]\]
\[ = \frac{1}{2}\left[ \sin \left( B - C + A - D \right) + \sin \left( B - C - A + D \right) \right] + \frac{1}{2}\left[ \sin \left\{ - \left( - C + A - B + D \right) \right\} + \sin \left\{ - \left( - C + A + B - D \right) \right\} \right] + \frac{1}{2}\left[ \sin\left\{ - \left( - A + B - C + D \right) \right\} + \sin \left( A - B - C + D \right) \right]\]
\[ = \frac{1}{2}\sin\left( B - C + A - D \right) + \frac{1}{2}\sin\left( B - C - A + D \right) - \frac{1}{2}\sin\left( - C + A - B + D \right) - \frac{1}{2}\sin\left( - C + A + B - D \right) - \frac{1}{2}\sin\left( - A + B - C + D \right) + \frac{1}{2}\sin\left( A - B - C + D \right)\]
\[ = 0\]
= RHS
APPEARS IN
संबंधित प्रश्न
Prove that:
Show that :
Prove that tan 20° tan 30° tan 40° tan 80° = 1.
Prove that:
sin 20° sin 40° sin 60° sin 80° = \[\frac{3}{16}\]
Prove that:
\[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right) = \tan 3x\]
Express each of the following as the product of sines and cosines:
sin 12x + sin 4x
Express each of the following as the product of sines and cosines:
sin 5x − sin x
Express each of the following as the product of sines and cosines:
cos 12x - cos 4x
Prove that:
cos 100° + cos 20° = cos 40°
Prove that:
cos 20° + cos 100° + cos 140° = 0
Prove that:
\[\sin\frac{5\pi}{18} - \cos\frac{4\pi}{9} = \sqrt{3} \sin\frac{\pi}{9}\]
Prove that:
Prove that:
`sin A + sin 2A + sin 4A + sin 5A = 4 cos (A/2) cos((3A)/2)sin3A`
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
If cos (α + β) sin (γ + δ) = cos (α − β) sin (γ − δ), prove that cot α cot β cot γ = cot δ
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), then write the value of tan A tan B tan C.
cos 35° + cos 85° + cos 155° =
sin 47° + sin 61° − sin 11° − sin 25° is equal to
If A, B, C are in A.P., then \[\frac{\sin A - \sin C}{\cos C - \cos A}\]=
If sin x + sin y = \[\sqrt{3}\] (cos y − cos x), then sin 3x + sin 3y =
If \[\tan\alpha = \frac{x}{x + 1}\] and
Express the following as the sum or difference of sine or cosine:
`sin "A"/8 sin (3"A")/8`
Express the following as the sum or difference of sine or cosine:
cos(60° + A) sin(120° + A)
Express the following as the sum or difference of sine or cosine:
`cos (7"A")/3 sin (5"A")/3`
Prove that:
cos 20° cos 40° cos 80° = `1/8`
Prove that:
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
If cos A + cos B = `1/2` and sin A + sin B = `1/4`, prove that tan `(("A + B")/2) = 1/2`
If cosec A + sec A = cosec B + sec B prove that cot`(("A + B"))/2` = tan A tan B.
If secx cos5x + 1 = 0, where 0 < x ≤ `pi/2`, then find the value of x.
