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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2024 · Mathematics · Trigonometric Ratios & Identities

What is 3 10^ ^ - 10^ ^ 25^ ^ 25^ ^ equal to?

What is \frac{\sqrt{3} \cos 10^{{}^{\circ}}-\sin 10^{{}^{\circ}}}{\sin 25^{{}^{\circ}} \cos 25^{{}^{\circ}}}$ equal to?

Options

  1. A1
  2. B3\sqrt{3}
  3. C2
  4. D4

Correct answer

D. 4

Step-by-step solution

3cos⁡10∘−sin⁡10∘sin⁡25∘⋅cos⁡25∘=(32⋅cos⁡10∘−12sin⁡10∘)22sin⁡25∘⋅cos⁡25∘2×2=4(sin⁡60∘⋅cos⁡10∘−cos⁡60∘⋅sin⁡10∘sin⁡50∘)=4⋅sin⁡(60∘−10∘)sin⁡50∘=4sin⁡50∘sin⁡50∘=4.\begin{aligned} & \frac{\sqrt{3} \cos 10^{{}^{\circ}}-\sin 10^{{}^{\circ}}}{\sin 25^{{}^{\circ}} \cdot \cos 25^{{}^{\circ}}} \\ &=\frac{\left(\frac{\sqrt{3}}{2} \cdot \cos 10^{{}^{\circ}}-\frac{1}{2} \sin 10^{{}^{\circ}}\right)}{2 \frac{2 \sin 25^{{}^{\circ}} \cdot \cos 25^{{}^{\circ}}}{2 \times 2}} \\ &= 4\left(\frac{\sin 60^{{}^{\circ}} \cdot \cos 10^{{}^{\circ}}-\cos 60^{{}^{\circ}} \cdot \sin 10^{{}^{\circ}}}{\sin 50^{{}^{\circ}}}\right) \\ &= \frac{4 \cdot \sin \left(60^{{}^{\circ}}-10^{{}^{\circ}}\right)}{\sin 50^{{}^{\circ}}} \\ &= \frac{4 \sin 50^{{}^{\circ}}}{\sin 50^{{}^{\circ}}}=4.\end{aligned}
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