Q80 / 150 · MHT CET 2020 (12 Oct Shift 1) · Mathematics

If $\frac{\sin (A+B)}{\sin (A-B)}=\frac{\cos (C+D)}{\cos (C-D)}$, then $\tan A \cot B=$

  • A
    $\cot C \cot D$
  • B
    $-\tan C \tan D$
  • C
    $\tan C \tan D$
  • Correct answer
    D
    $-\cot C \cot D$

Solution

$\frac{\sin (A+B)}{\sin (A-B)}=\frac{\cos (C+D)}{\cos (C-D)}$ $\therefore \frac{\sin (A+B)+\sin (A-B)}{\sin (A+B)-\sin (A-B)}=\frac{\cos (C+D)+\cos (C-D)}{\cos (C+D)-\cos (C-D)}$ $\therefore \frac{2 \sin A \cos B}{2 \cos A \sin B}=\frac{2 \cos C \cos D}{-2 \sin C \sin D}$ $\quad \tan A \cot B=-\cot C \cot D$

Hence, the correct answer is option: 'D'.

Relevant Questions