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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2020 · Mathematics · Trigonometric Ratios & Identities

What is the value of 8 10^ ^ · 20^ ^ · 40^ ^ ?

What is the value of 8cos⁡10∘⋅cos⁡20∘⋅cos⁡40∘8 \cos 10^{{}^{\circ}} \cdot \cos 20^{{}^{\circ}} \cdot \cos 40^{{}^{\circ}} ?

Options

  1. Atan⁡10∘\tan 10^{{}^{\circ}}
  2. Bcot⁡10∘\cot 10^{{}^{\circ}}
  3. Ccosec⁡10∘\operatorname{cosec} 10^{{}^{\circ}}
  4. Dsec⁡10∘\sec 10^{{}^{\circ}}

Correct answer

B. cot⁡10∘\cot 10^{{}^{\circ}}

Step-by-step solution

8⋅cos⁡10∘⋅cos⁡20∘⋅cos⁡40∘=4sin⁡10∘[2⋅cos⁡10∘⋅sin⁡10∘]⋅cos⁡20∘⋅cos⁡40∘=4sin⁡10∘[sin⁡20∘]⋅cos⁡20∘⋅cos⁡40∘=2sin⁡10∘[2sin⁡20∘⋅cos⁡20∘]⋅cos⁡40∘\begin{aligned} & 8 \cdot \cos 10^{{}^{\circ}} \cdot \cos 20^{{}^{\circ}} \cdot \cos 40^{{}^{\circ}} \\ &= \frac{4}{\sin 10^{{}^{\circ}}}\left[2 \cdot \cos 10^{{}^{\circ}} \cdot \sin 10^{{}^{\circ}}\right] \cdot \cos 20^{{}^{\circ}} \cdot \cos 40^{{}^{\circ}} \\ &= \frac{4}{\sin 10^{{}^{\circ}}}\left[\sin 20^{{}^{\circ}}\right] \cdot \cos 20^{{}^{\circ}} \cdot \cos 40^{{}^{\circ}} \\ &= \frac{2}{\sin 10^{{}^{\circ}}}\left[2 \sin 20^{{}^{\circ}} \cdot \cos 20^{{}^{\circ}}\right] \cdot \cos 40^{{}^{\circ}}\end{aligned}=2sin⁡10∘[sin⁡40∘⋅j0cos⁡40∘=1sin⁡10∘[2sin⁡40∘⋅cos⁡40∘]=sin⁡80∘sin⁡10∘=sin⁡(90∘−10∘)sin⁡10∘=cos⁡10∘sin⁡10∘=cot⁡10∘\begin{aligned} & =\frac{2}{\sin 10^{{}^{\circ}}}\left[\sin 40^{{}^{\circ}} \cdot j 0 \cos 40^{{}^{\circ}}\right. \\ & =\frac{1}{\sin 10^{{}^{\circ}}}\left[2 \sin 40^{{}^{\circ}} \cdot \cos 40^{{}^{\circ}}\right]=\frac{\sin 80^{{}^{\circ}}}{\sin 10^{{}^{\circ}}} \\ & =\frac{\sin \left(90^{{}^{\circ}}-10^{{}^{\circ}}\right)}{\sin 10^{{}^{\circ}}}=\frac{\cos 10^{{}^{\circ}}}{\sin 10^{{}^{\circ}}}=\cot 10^{{}^{\circ}}\end{aligned}$
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