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Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles1 Mark
MCQ
If $\text{A+B+C}=\pi,$ then $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$ is equal to:
  • A
    $\tan\text{A}\tan\text{B}\tan\text{C}$
  • B
    $0$
  • ✓
    $1$
  • D
    None of these
✓

Answer

Correct option: C.
$1$
$\pi=180^\circ$
Using $\tan(180^\circ-\text{A})=-\tan\text{A},$ we get:
$\text{C}=\pi-(\text{A+B})$
Now, $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan[\pi-\text{(A+B)}]}{\tan\text{A}\tan\text{B}\tan[\pi-\text{(A+B)}]}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan\text{(A+B)}}{-\tan\text{A}\tan\text{B}\tan\text{(A+B)}}$
$=\frac{\tan\text{A}+\tan\text{B}-\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}}{-\tan\text{A}\tan\text{B}\times\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}-\tan\text{A}-\tan\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$
$=\frac{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$
$=1$

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