If A+B+C=360∘ then cot⁡A4+cot⁡B4+cot⁡C4=

A+B+C=3600A4+B4+C4=90∘⇒A4+B4=90∘−C4⇒cot⁡A4+B4=cot⁡90−C4⇒cot⁡A4cot⁡B4−1cot⁡B4+cot⁡A4=1cot⁡C4⇒cot⁡A4cot⁡B4cot⁡C4−cot⁡C4=cot⁡A4+cot⁡B4∴cot⁡A4+cot⁡B4+cot⁡C4=cot⁡A4cot⁡B4cot⁡C4

If A+B+C=360∘ then cot⁡A4+cot⁡B4+cot⁡C4=

A+B+C=3600A4+B4+C4=90∘⇒A4+B4=90∘−C4⇒cot⁡A4+B4=cot⁡90−C4⇒cot⁡A4cot⁡B4−1cot⁡B4+cot⁡A4=1cot⁡C4⇒cot⁡A4cot⁡B4cot⁡C4−cot⁡C4=cot⁡A4+cot⁡B4∴cot⁡A4+cot⁡B4+cot⁡C4=cot⁡A4cot⁡B4cot⁡C4

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If $A + B + C = 360^{\circ}$ then $\cot \frac{A}{4} + \cot \frac{B}{4} + \cot \frac{C}{4} =$

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$\cos \frac{A}{4} \times \cos \frac{B}{4} \times \cos \frac{C}{4}$

$\cot \frac{A}{4} \times \cot \frac{B}{4} \times \cot \frac{C}{4}$

$\tan \frac{A}{4} \times \tan \frac{B}{4} \times \tan \frac{C}{4}$

$\sin \frac{A}{4} \times \sin \frac{B}{4} \times \sin \frac{C}{4}$

answer is D.

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Detailed Solution

$A + B + C = 360^{0}$

$\frac{A}{4} + \frac{B}{4} + \frac{C}{4} = 90^{\circ} \Rightarrow \frac{A}{4} + \frac{B}{4} = 90^{\circ} - \frac{C}{4}$

$\Rightarrow \cot (\frac{A}{4} + \frac{B}{4}) = \cot (90 - \frac{C}{4})$

$\Rightarrow \frac{\cot \frac{A}{4} \cot \frac{B}{4} - 1}{\cot \frac{B}{4} + \cot \frac{A}{4}} = \frac{1}{\cot \frac{C}{4}} \Rightarrow \cot \frac{A}{4} \cot \frac{B}{4} \cot \frac{C}{4} - \cot \frac{C}{4} = \cot \frac{A}{4} + \cot \frac{B}{4}$