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
Show that :
Prove that:
cos 40° cos 80° cos 160° = \[- \frac{1}{8}\]
Prove that:
sin 20° sin 40° sin 60° sin 80° = \[\frac{3}{16}\]
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 12x + sin 4x
Prove that:
sin 38° + sin 22° = sin 82°
Prove that:
cos 55° + cos 65° + cos 175° = 0
Prove that:
Prove that:
Prove that:
sin 3A + sin 2A − sin A = 4 sin A cos \[\frac{A}{2}\] \[\frac{3A}{2}\]
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
Prove that:
If cosec A + sec A = cosec B + sec B, prove that tan A tan B = \[\cot\frac{A + B}{2}\].
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), prove that tan A tan B tan C + tan D = 0.
If A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].
If sin 2A = λ sin 2B, then write the value of \[\frac{\lambda + 1}{\lambda - 1}\]
If sin 2 θ + sin 2 ϕ = \[\frac{1}{2}\] and cos 2 θ + cos 2 ϕ = \[\frac{3}{2}\], then cos2 (θ − ϕ) =
cos 35° + cos 85° + cos 155° =
If cos A = m cos B, then \[\cot\frac{A + B}{2} \cot\frac{B - A}{2}\]=
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 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-
cos 20° + cos 100° + cos 140°
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:
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)]`
