Advertisements
Advertisements
Question
Advertisements
Solution
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
RELATED QUESTIONS
Show that :
Prove that:
tan 20° tan 40° tan 60° tan 80° = 3
Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0
Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
If α + β = \[\frac{\pi}{2}\], show that the maximum value of cos α cos β is \[\frac{1}{2}\].
Express each of the following as the product of sines and cosines:
sin 5x − sin x
Prove that:
sin 105° + cos 105° = cos 45°
Prove that:
sin 50° − sin 70° + sin 10° = 0
Prove that:
cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
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:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
cos (A + B + C) + cos (A − B + C) + cos (A + B − C) + cos (− A + B + C) = 4 cos A cos Bcos C
Prove that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
Write the value of sin \[\frac{\pi}{12}\] sin \[\frac{5\pi}{12}\].
If sin A + sin B = α and cos A + cos B = β, then write the value of tan \[\left( \frac{A + B}{2} \right)\].
If A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].
Write the value of \[\frac{\sin A + \sin 3A}{\cos A + \cos 3A}\]
cos 40° + cos 80° + cos 160° + cos 240° =
If sin 2 θ + sin 2 ϕ = \[\frac{1}{2}\] and cos 2 θ + cos 2 ϕ = \[\frac{3}{2}\], then cos2 (θ − ϕ) =
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:
cos(60° + A) sin(120° + A)
Express the following as the product of sine and cosine.
sin A + sin 2A
Express the following as the product of sine and cosine.
cos 2θ – cos θ
Prove that:
`(cos 2"A" - cos 3"A")/(sin "2A" + sin "3A") = tan "A"/2`
Prove that:
`(cos 7"A" +cos 5"A")/(sin 7"A" −sin 5"A")` = cot A
Prove that cos 20° cos 40° cos 60° cos 80° = `3/16`.
Find the value of tan22°30′. `["Hint:" "Let" θ = 45°, "use" tan theta/2 = (sin theta/2)/(cos theta/2) = (2sin theta/2 cos theta/2)/(2cos^2 theta/2) = sintheta/(1 + costheta)]`
