If (A+B) (C-D)= (A-B) (C+D), prove that + =0

Question

$\text{If}\ \cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D}),$
prove that $\tan\text{A}\tan\text{B}\tan\text{C}+\tan\text{D}=0$

✓

Answer

We have,
$\cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D})$
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}...(\text{i})$
Now,
$\frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}$
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}+1=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}+1$
$\Rightarrow\ \frac{\cos(\text{A+B})+\cos(\text{A}-\text{B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})+\sin(\text{C}-\text{D})}{\sin(\text{C}-\text{D})}...(\text{ii})$
Again,
$\frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}$ [By equation (i)]
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}-1=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}-1$
$\Rightarrow\ \frac{\cos(\text{A+B})-\cos(\text{A}-\text{B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})-\sin(\text{C}-\text{D})}{\sin(\text{C}-\text{D})}...(\text{iii})$
Dividing equation (ii) by equation (iii), we get
$\frac{\cos(\text{A+B})+\cos(\text{A}-\text{B})}{\cos(\text{A+B})-\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})+\sin(\text{C}-\text{D})}{\sin(\text{C+D})-\sin(\text{C}-\text{D})}$
$\Rightarrow\ \frac{2\cos\big\{\frac{\text{A+B+A}-\text{B}}{2}\big\}\cos\big\{\frac{\text{A+B}-\text{A+B}}{2}\big\}}{-2\sin\big\{\frac{\text{A+B+A}-\text{B}}{2}\big\}\sin\big\{\frac{\text{A+B}-\text{A+B}}{2}\big\}}\\ \ \ \ =\frac{2\sin\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}\cos\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}}{2\sin\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}\cos\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}}$
$\Rightarrow\ \frac{\cos\text{A}\cos\text{B}}{-\sin\text{A}\sin\text{B}}=\frac{\sin\text{C}\cos\text{D}}{\sin\text{D}\cos\text{C}}$
$\Rightarrow\ \frac{1}{-\tan\text{A}\tan\text{B}}=\frac{\sin\text{C}\cos\text{D}}{\cos\text{C}\sin\text{D}}$
$\Rightarrow\ \frac{-1}{\tan\text{A}\tan\text{B}}=\frac{\tan\text{C}}{\tan\text{D}}$
$\Rightarrow\ -\tan\text{D}=\tan\text{A}\tan\text{B}\tan\text{C}$
$\Rightarrow\ \tan\text{A}\tan\text{B}\tan\text{C}=-\tan\text{D}$
$\Rightarrow\ \tan\text{A}\tan\text{B}\tan\text{C}+\tan\text{D}=0$ Hence proved.

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Match the following sets for all sets A, B and C.

(i)	$((\text{A}'\cup\text{B}')-\text{A})'$	(a)	$\text{A} - \text{B}$
(ii)	$[\text{B}'\cup(\text{B}'-\text{A})]'$	(b)	$\text{A}$
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