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

NDA 2024 · Mathematics · Trigonometric Ratios & Identities

What is the value of 10^ ^ · 50^ ^ + 50^ ^ . 250^ ^ + 250^ ^ · 10^...

What is the value of sin⁡10∘⋅sin⁡50∘+sin⁡50∘\sin 10^{{}^{\circ}} \cdot \sin 50^{{}^{\circ}}+\sin 50^{{}^{\circ}}. sin⁡250∘+sin⁡250∘⋅sin⁡10∘\sin 250^{{}^{\circ}}+\sin 250^{{}^{\circ}} \cdot \sin 10^{{}^{\circ}} equal to?

Options

  1. A−14-\frac{1}{4}
  2. B−34-\frac{3}{4}
  3. C3sin⁡10∘4\frac{3 \sin 10^{{}^{\circ}}}{4}
  4. D−3cos⁡10∘4-\frac{3 \cos 10^{{}^{\circ}}}{4}

Correct answer

B. −34-\frac{3}{4}

Step-by-step solution

sin⁡10∘⋅sin⁡50∘+sin⁡50∘⋅sin⁡250∘+sin⁡250∘⋅sin⁡10∘\begin{aligned} & \sin 10^{{}^{\circ}} \cdot \sin 50^{{}^{\circ}}+ \sin 50^{{}^{\circ}} \cdot \sin 250^{{}^{\circ}} \\ &+\sin 250^{{}^{\circ}} \cdot \sin 10^{{}^{\circ}}\end{aligned}=12(cos⁡40∘−cos⁡60∘+cos⁡200∘−cos⁡300∘+cos⁡240∘−cos⁡260∘=12[cos⁡40∘−12+cos⁡(180+20)∘−cos⁡(360−60)∘+cos⁡(180+60)∘−cos⁡(180+80)∘]=12[cos⁡40∘−12−cos⁡20∘−cos⁡60∘−cos⁡60∘+cos⁡80∘]=12[cos⁡40∘−cos⁡20∘+cos⁡80∘−12−12−12]\begin{aligned} & =\frac{1}{2}\left(\cos 40^{{}^{\circ}}-\cos 60^{{}^{\circ}}+\cos 200^{{}^{\circ}}\right. \\ & \quad-\cos 300^{{}^{\circ}}+\cos 240^{{}^{\circ}}-\cos 260^{{}^{\circ}} \\ & =\frac{1}{2}\left[\begin{array}{l}\cos 40^{{}^{\circ}}-\frac{1}{2}+\cos (180+20)^{{}^{\circ}}-\cos (360-60)^{{}^{\circ}} \\ +\cos (180+60)^{{}^{\circ}}-\cos (180+80)^{{}^{\circ}}\end{array}\right] \\ & =\frac{1}{2}\left[\cos 40^{{}^{\circ}}-\frac{1}{2}-\cos 20^{{}^{\circ}}-\cos 60^{{}^{\circ}}-\cos 60^{{}^{\circ}}+\cos 80^{{}^{\circ}}\right] \\ & =\frac{1}{2}\left[\cos 40^{{}^{\circ}}-\cos 20^{{}^{\circ}}+\cos 80^{{}^{\circ}}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right]\end{aligned}=12[cos⁡40∘+cos⁡80∘−cos⁡20∘−32]=12[2cos⁡(80+402)×cos⁡(80−402)−cos⁡20∘−32]=12[2cos⁡60∘⋅cos⁡20∘−cos⁡20∘−32]=12[2×12⋅cos⁡20∘−cos⁡20∘−32]=12[cos⁡20∘−cos⁡20∘−32]=12[−32]=−34\begin{aligned} & =\frac{1}{2}\left[\cos 40^{{}^{\circ}}+\cos 80^{{}^{\circ}}-\cos 20^{{}^{\circ}}-\frac{3}{2}\right] \\ & =\frac{1}{2}\left[2 \cos \left(\frac{80+40}{2}\right) \times \cos \left(\frac{80-40}{2}\right)-\cos 20^{{}^{\circ}}-\frac{3}{2}\right] \\ & =\frac{1}{2}\left[2 \cos 60^{{}^{\circ}} \cdot \cos 20^{{}^{\circ}}-\cos 20^{{}^{\circ}}-\frac{3}{2}\right] \\ & =\frac{1}{2}\left[2 \times \frac{1}{2} \cdot \cos 20^{{}^{\circ}}-\cos 20^{{}^{\circ}}-\frac{3}{2}\right] \\ & =\frac{1}{2}\left[\cos 20^{{}^{\circ}}-\cos 20^{{}^{\circ}}-\frac{3}{2}\right] \\ & =\frac{1}{2}\left[-\frac{3}{2}\right]=-\frac{3}{4}\end{aligned}$
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