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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2025 · Mathematics · Inverse Trigonometric Functions

What is the value of [(1)/(2) ^ -1 ( 2 3 )]?

What is the value of tan⁡\tan\left[\dfrac{1}{2}\sec^{-1} \left(\dfrac{2}{\sqrt{3}}\right)\right]?

Options

  1. A2−32 - \sqrt{3}
  2. B2+32 + \sqrt{3}
  3. C3−1\sqrt{3} - 1
  4. D3+1\sqrt{3} + 1

Correct answer

A. 2−32 - \sqrt{3}

Step-by-step solution

Let θ=sec⁡−1(23)<br><br>⇒sec⁡θ=23<br><br>⇒θ=π6<br><br>Thegivenexpressionbecomestan⁡(12θ)=tan⁡(π12)<br><br>tan⁡(π12)=tan⁡(15∘)=tan⁡(45∘−30∘)\theta = \sec^{-1} \left(\dfrac{2}{\sqrt{3}}\right)<br><br>\Rightarrow \sec \theta = \dfrac{2}{\sqrt{3}}<br><br> ⇒ \theta = \dfrac{\pi}{6}<br><br>The given expression becomes \tan \left(\dfrac{1}{2} \theta\right) = \tan \left(\dfrac{\pi}{12}\right)<br><br>\tan\left(\dfrac{\pi}{12}\right) = \tan(15^{{}^{\circ}}) = \tan(45^{{}^{\circ}} - 30^{{}^{\circ}})

Using the formula tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B<br><br>tan⁡(15∘)=tan⁡45∘−tan⁡30∘1+tan⁡45∘tan⁡30∘<br><br>tan⁡(15∘)=1−131+13=3−13+1<br><br>Rationalizingthedenominator:<br><br>tan⁡(15∘)=(3−1)2(3+1)(3−1)=3+1−233−1<br><br>tan⁡(15∘)=4−232=2−3<br><br>Answer:2−3\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}<br><br>\tan(15^{{}^{\circ}}) = \dfrac{\tan 45^{{}^{\circ}} - \tan 30^{{}^{\circ}}}{1 + \tan 45^{{}^{\circ}} \tan 30^{{}^{\circ}}}<br><br>\tan(15^{{}^{\circ}}) = \dfrac{1 - \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{\sqrt{3}}} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}<br><br>Rationalizing the denominator:<br><br>\tan(15^{{}^{\circ}}) = \dfrac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \dfrac{3 + 1 - 2\sqrt{3}}{3 - 1}<br><br>\tan(15^{{}^{\circ}}) = \dfrac{4 - 2\sqrt{3}}{2} =2 - \sqrt{3}<br><br>Answer: 2 - \sqrt{3}
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