MHT CET · Maths · Trigonometric Equations
If \(\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}\), then \(\tan A \tan B \tan C=\)
- A 0
- B \(\tan \mathrm{D}\)
- C \(\cot D\)
- D \(-\tan D\)
Answer & Solution
Correct Answer
(D) \(-\tan D\)
Step-by-step Solution
Detailed explanation
\(\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}\)
By Componendo - Dividendo, we write
\(\frac{\cos (A+B)+\cos (A-B)}{\cos (A+B)-\cos (A-B)}=\frac{\sin (C+D)+\sin (C-D)}{\sin (C+D)-\sin (C-D)} \)
\( \frac{2 \cos A \cos B}{-2 \sin A \sin B}=\frac{2 \sin C \cos D}{2 \cos C \sin D} \)
\( \therefore \frac{1}{-\tan A \tan B}=\frac{\tan C}{\tan D} \Rightarrow \tan A \tan B \tan C\)\(=-\tan D\)
By Componendo - Dividendo, we write
\(\frac{\cos (A+B)+\cos (A-B)}{\cos (A+B)-\cos (A-B)}=\frac{\sin (C+D)+\sin (C-D)}{\sin (C+D)-\sin (C-D)} \)
\( \frac{2 \cos A \cos B}{-2 \sin A \sin B}=\frac{2 \sin C \cos D}{2 \cos C \sin D} \)
\( \therefore \frac{1}{-\tan A \tan B}=\frac{\tan C}{\tan D} \Rightarrow \tan A \tan B \tan C\)\(=-\tan D\)
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