Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[\frac{\cos \left( A - B \right)}{\cos \left( A + B \right)} + \frac{\cos \left( C + D \right)}{\cos \left( C - D \right)} = 0\]
\[ \Rightarrow \frac{\cos \left( A - B \right) \cos \left( C - D \right) + \cos \left( C + D \right) \cos \left( A + B \right)}{\cos \left( A + B \right) \cos \left( C - D \right)} = 0\]
\[ \Rightarrow \cos \left( A - B \right) \cos \left( C - D \right) + \cos \left( C + D \right) \cos \left( A + B \right) = 0\]
\[ \Rightarrow \cos \left( A - B \right) \cos \left( C - D \right) = - \cos \left( C + D \right) \cos \left( A + B \right)\]
\[ \Rightarrow \left[ \cos A \cos B + \sin A \sin B \right]\left[ \cos C \cos D + \sin C \sin D \right] = - \left[ \cos C \cos D - \sin C \sin D \right]\left[ \cos A \cos B - \sin A \sin B \right]\]
\[\text{ Dividing both sides by }\cos A \cos B \cos C \cos D \text{ we get, }\]
\[\frac{\left[ \cos A \cos B + \sin A\sin B \right]\left[ \cos C\cos D + \sin C\sin D \right]}{\cos A\cos B\cos C\cos D} = - \frac{\left[ \cos C \cos D - \sin C\sin D \right]\left[ \cos A\cos B - \sin A\sin B \right]}{\cos A\cos B\cos C\cos D}\]
\[ \Rightarrow \frac{\left[ \cos A \cos B + \sin A\sin B \right]}{\cos A\cos B} \times \frac{\left[ \cos C\cos D + \sin C\sin D \right]}{\cos C\cos D} = - \frac{\left[ \cos C \cos D - \sin C\sin D \right]}{\cos C\cos D} \times \frac{\left[ \sin C\cos A\cos B - \sin A\sin B \right]}{\cos A\cos B}\]
\[ \Rightarrow \left[ 1 + \tan A\tan B \right]\left[ 1 + \tan C\tan D \right] = \left[ \tan C\tan D - 1 \right]\left[ 1 - \tan A\tan B \right]\]
\[ \Rightarrow 1 + \tan C\tan D + \tan A\tan B + \tan A\tan B\tan C\tan D = \tan C\tan D - \tan A\tan B\tan C\tan D + \tan A\tan B\tan D - 1 + \tan A\tan B\]
\[ \Rightarrow 2\tan A\tan B\tan C\tan D = - 2\]
\[ \Rightarrow \tan A\tan B\tan C\tan D = - 1\]
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove that:
cos 40° cos 80° cos 160° = \[- \frac{1}{8}\]
Prove that:
sin 10° sin 50° sin 60° sin 70° = \[\frac{\sqrt{3}}{16}\]
Prove that:
sin 20° sin 40° sin 60° sin 80° = \[\frac{3}{16}\]
Express each of the following as the product of sines and cosines:
cos 12x + cos 8x
Express each of the following as the product of sines and cosines:
sin 2x + cos 4x
Prove that:
sin 23° + sin 37° = cos 7°
Prove that:
cos 20° + cos 100° + cos 140° = 0
Prove that:
cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
Prove that:
`sin A + sin 2A + sin 4A + sin 5A = 4 cos (A/2) cos((3A)/2)sin3A`
Prove that:
cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −\[\frac{3}{4}\]
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), prove that tan A tan B tan C + tan D = 0.
Write the value of the expression \[\frac{1 - 4 \sin 10^\circ \sin 70^\circ}{2 \sin 10^\circ}\]
If A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].
cos 40° + cos 80° + cos 160° + cos 240° =
sin 163° cos 347° + sin 73° sin 167° =
The value of sin 78° − sin 66° − sin 42° + sin 60° is ______.
cos 35° + cos 85° + cos 155° =
sin 47° + sin 61° − sin 11° − sin 25° 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
If sin x + sin y = \[\sqrt{3}\] (cos y − cos x), then sin 3x + sin 3y =
Express the following as the sum or difference of sine or cosine:
`sin "A"/8 sin (3"A")/8`
Express the following as the product of sine and cosine.
cos 2θ – cos θ
Prove that:
2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\] = 0
Prove that:
`(cos 2"A" - cos 3"A")/(sin "2A" + sin "3A") = tan "A"/2`
Prove that cos 20° cos 40° cos 60° cos 80° = `3/16`.
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.
