मराठी

Show That: Sin (B − C) Cos (A − D) + Sin (C − A) Cos (B − D) + Sin (A − B) Cos (C − D) = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0

बेरीज
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

shaalaa.com
Transformation Formulae
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Transformation formulae - Exercise 8.1 [पृष्ठ ७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 8 Transformation formulae
Exercise 8.1 | Q 6.2 | पृष्ठ ७

संबंधित प्रश्‍न

Prove that:

\[2\sin\frac{5\pi}{12}\sin\frac{\pi}{12} = \frac{1}{2}\]

 


Prove that: 

\[2\sin\frac{5\pi}{12}\cos\frac{\pi}{12} = \frac{\sqrt{3} + 2}{2}\]

\[\text{ Prove that }4 \cos x \cos\left( \frac{\pi}{3} + x \right) \cos \left( \frac{\pi}{3} - x \right) = \cos 3x .\]

 


Prove that:
cos 10° cos 30° cos 50° cos 70° = \[\frac{3}{16}\]

 


Prove that:
 sin 20° sin 40° sin 80° = \[\frac{\sqrt{3}}{8}\]

 


Prove that:
cos 20° cos 40° cos 80° = \[\frac{1}{8}\]

 


Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0


Express each of the following as the product of sines and cosines:
 cos 12x + cos 8x


Prove that:
 sin 23° + sin 37° = cos 7°


Prove that:
sin 105° + cos 105° = cos 45°


Prove that:
 cos 80° + cos 40° − cos 20° = 0


Prove that:
\[\sin\frac{5\pi}{18} - \cos\frac{4\pi}{9} = \sqrt{3} \sin\frac{\pi}{9}\]


Prove that:

\[\cos\left( \frac{3\pi}{4} + x \right) - \cos\left( \frac{3\pi}{4} - x \right) = - \sqrt{2} \sin x\]

 


Prove that:

\[\cos\left( \frac{\pi}{4} + x \right) + \cos\left( \frac{\pi}{4} - x \right) = \sqrt{2} \cos x\]

 


Prove that:

\[\frac{\sin A + \sin B}{\sin A - \sin B} = \tan \left( \frac{A + B}{2} \right) \cot \left( \frac{A - B}{2} \right)\]

Prove that:

\[\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A} = \tan 3A\]

 


Prove that:

\[\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A} = \cot 6A\]

Prove that:

\[\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A} = \tan A\]

Prove that:
cos (A + B + C) + cos (A − B + C) + cos (A + B − C) + cos (− A + B + C) = 4 cos A cos Bcos C


If y sin ϕ = x sin (2θ + ϕ), prove that (x + y) cot (θ + ϕ) = (y − x) cot θ.

 

Write the value of sin \[\frac{\pi}{12}\] sin \[\frac{5\pi}{12}\].


Write the value of the expression \[\frac{1 - 4 \sin 10^\circ \sin 70^\circ}{2 \sin 10^\circ}\]


Write the value of \[\sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10}\]


Write the value of \[\frac{\sin A + \sin 3A}{\cos A + \cos 3A}\]


cos 40° + cos 80° + cos 160° + cos 240° =


The value of sin 78° − sin 66° − sin 42° + sin 60° is ______.


The value of sin 50° − sin 70° + sin 10° is equal to


If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in


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  (7"A")/3 sin  (5"A")/3`


Prove that:

(cos α – cos β)2 + (sin α – sin β)2 = 4 sin2 `((alpha - beta)/2)`


Evaluate:

sin 50° – sin 70° + sin 10°


If cos A + cos B = `1/2` and sin A + sin B = `1/4`, prove that tan `(("A + B")/2) = 1/2`


If sin(y + z – x), sin(z + x – y), sin(x + y – z) are in A.P, then prove that tan x, tan y and tan z are in A.P.


If tan θ = `1/sqrt5` and θ lies in the first quadrant then cos θ is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×