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