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Question: Answered & Verified by Expert
In $\triangle \mathrm{ABC},(\cot \mathrm{A}+\cot \mathrm{B})(\cot \mathrm{B}+\cot \mathrm{C})(\cot \mathrm{C}+\cot \mathrm{A})=$
MathematicsTrigonometric Ratios & IdentitiesAP EAMCETAP EAMCET 2023 (16 May Shift 2)
Options:
  • A $\sec A \sec B \sec C$
  • B $\tan \mathrm{A} \tan \mathrm{B} \tan \mathrm{C}$
  • C $\operatorname{cosec} A \operatorname{cosec} B \operatorname{cosec} C$
  • D $\cot A \cot B \cot C$
Solution:
2757 Upvotes Verified Answer
The correct answer is: $\operatorname{cosec} A \operatorname{cosec} B \operatorname{cosec} C$
Since $(\cot \mathrm{A}+\cot \mathrm{B})(\cot \mathrm{B}+\cot \mathrm{C})(\cot \mathrm{C}+\cot \mathrm{A})$
$\begin{aligned}
& =\left(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}\right)\left(\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}\right)\left(\frac{\cos C}{\sin C}+\frac{\cos A}{\sin A}\right) \\
& =\left(\frac{\cos A \cdot \sin B+\cos B \sin A}{\sin A \cdot \sin B}\right)\left(\frac{\cos B \sin C+\cos C \sin B}{\sin B \cdot \sin C}\right) \times \\
& \left(\frac{\cos C \sin A+\cos A \sin C}{\sin C \cdot \sin A}\right)
\end{aligned}$
$\begin{aligned} & =\frac{\sin (A+B)}{\sin A \sin B} \times \frac{\sin (B+C)}{\sin B \sin C} \times \frac{\sin (A+C)}{\sin A \sin C} \\ & =\frac{\sin (\pi-C)}{\sin A \cdot \sin B} \times \frac{\sin (\pi-A)}{\sin B \sin C} \times \frac{\sin (\pi-B)}{\sin A \cdot \sin C} \\ & =\frac{\sin C}{\sin A \cdot \sin B} \times \frac{\sin A}{\sin B \sin C} \times \frac{\sin B}{\sin A \sin C} \\ & =\operatorname{cosec} A \cdot \operatorname{cosec} B \cdot \operatorname{cosec} C\end{aligned}$

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