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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

What is 25^ ^ 15^ ^ + 15^ ^ 50^ ^ + 25^ ^ 50^ ^ equal to?

What is tan⁡25∘tan⁡15∘+tan⁡15∘tan⁡50∘+tan⁡25∘tan⁡50∘\tan 25^{{}^{\circ}} \tan 15^{{}^{\circ}}+\tan 15^{{}^{\circ}} \tan 50^{{}^{\circ}}+\tan 25^{{}^{\circ}} \tan 50^{{}^{\circ}} equal to?

Options

  1. A0
  2. B1
  3. C2
  4. D4

Correct answer

B. 1

Step-by-step solution

tan⁡25∘⋅tan⁡15∘+tan⁡15∘⋅tan⁡50∘+tan⁡50∘⋅tan⁡25∘=tan⁡15∘(tan⁡15∘+tan⁡50∘)+tan⁡50∘⋅tan⁡25∘⋯(1)∵tan⁡75∘=tan⁡25∘+tan⁡50∘1−tan⁡25∘⋅tan⁡50∘=cot⁡15∘⇒tan⁡25∘+tan⁡50∘=cot⁡15∘(1−tan⁡25∘⋅tan⁡50∘) Put in  (1)⇒tan⁡15∘⋅cot⁡15∘⋅(1−tan⁡25∘⋅tan⁡50∘)=1−tan⁡25∘⋅tan⁡50∘+tan⁡50∘⋅tan⁡25∘+tan⁡50∘⋅tan⁡25∘=1\begin{aligned} & \tan 25^{{}^{\circ}} \cdot \tan 15^{{}^{\circ}}+\tan 15^{{}^{\circ}} \cdot \tan 50^{{}^{\circ}}+\tan 50^{{}^{\circ}} \cdot \tan 25^{{}^{\circ}} \\ & =\tan 15^{{}^{\circ}}\left(\tan 15^{{}^{\circ}}+\tan 50^{{}^{\circ}}\right)+\tan 50^{{}^{\circ}} \cdot \tan 25^{{}^{\circ}} \\ & \cdots(1) \\ & \because \tan 75^{{}^{\circ}}=\frac{\tan 25^{{}^{\circ}}+\tan 50^{{}^{\circ}}}{1-\tan 25^{{}^{\circ}} \cdot \tan 50^{{}^{\circ}}}=\cot 15^{{}^{\circ}} \\ & \Rightarrow \tan 25^{{}^{\circ}}+\tan 50^{{}^{\circ}}=\cot 15^{{}^{\circ}}\left(1-\tan 25^{{}^{\circ}} \cdot \tan 50^{{}^{\circ}}\right) \\ & \text{ Put\text{ in } }(1) \\ & \Rightarrow \tan 15^{{}^{\circ}} \cdot \cot 15^{{}^{\circ}} \cdot\left(1-\tan 25^{{}^{\circ}} \cdot \tan 50^{{}^{\circ}}\right) \\ & =1-\tan 25^{{}^{\circ}} \cdot \tan 50^{{}^{\circ}}+\tan 50^{{}^{\circ}} \cdot \tan 25^{{}^{\circ}}+\tan 50^{{}^{\circ}} \cdot \tan 25^{{}^{\circ}} \\ & =1 \end{aligned}
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