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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2017 · Mathematics · Trigonometric Ratios & Identities

The value of 3 cosec 20^ ^ - 20^ ^ is equal to

The value of \sqrt{3} \operatorname{cosec} 20^{{}^{\circ}}-\sec 20^{{}^{\circ}} is equal to

Options

  1. A44
  2. B22
  3. C11
  4. D−4-4

Correct answer

A. 44

Step-by-step solution

3cosec⁡20∘−sec⁡20∘=3sin⁡20∘−1cos⁡20∘=3cos⁡20∘−sin⁡20∘sin⁡20∘cos⁡20∘\begin{aligned} & \sqrt{3} \operatorname{cosec} 20^{{}^{\circ}}-\sec 20^{{}^{\circ}}=\frac{\sqrt{3}}{\sin 20^{{}^{\circ}}}-\frac{1}{\cos 20^{{}^{\circ}}} \\ & =\frac{\sqrt{3} \cos 20^{{}^{\circ}}-\sin 20^{{}^{\circ}}}{\sin 20^{{}^{\circ}} \cos 20^{{}^{\circ}}}\end{aligned}=2(32cos⁡20∘−12sin⁡20∘)sin⁡20∘cos⁡20∘=2(cos⁡30∘cos⁡20∘−sin⁡30∘sin⁡20∘)(2sin⁡20∘cos⁡20∘2)\begin{aligned} & =\frac{2\left(\frac{\sqrt{3}}{2} \cos 20^{{}^{\circ}}-\frac{1}{2} \sin 20^{{}^{\circ}}\right)}{\sin 20^{{}^{\circ}} \cos 20^{{}^{\circ}}} \\ & =\frac{2\left(\cos 30^{{}^{\circ}} \cos 20^{{}^{\circ}}-\sin 30^{{}^{\circ}} \sin 20^{{}^{\circ}}\right)}{\left(\frac{2 \sin 20^{{}^{\circ}} \cos 20^{{}^{\circ}}}{2}\right)}\end{aligned}=4[cos⁡(30∘+20∘)]sin⁡40∘=4cos⁡50∘sin⁡40∘=4cos⁡(90∘−40∘)sin⁡40∘=4sin⁡40∘sin⁡40∘=4.\begin{aligned} & =\frac{4\left[\cos \left(30^{{}^{\circ}}+20^{{}^{\circ}}\right)\right]}{\sin 40^{{}^{\circ}}} \\ & =\frac{4 \cos 50^{{}^{\circ}}}{\sin 40^{{}^{\circ}}}=\frac{4 \cos \left(90^{{}^{\circ}}-40^{{}^{\circ}}\right)}{\sin 40^{{}^{\circ}}}=\frac{4 \sin 40^{{}^{\circ}}}{\sin 40^{{}^{\circ}}}=4 .\end{aligned}
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