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UPSC NDA I 2017 Mathematics Solved Paper

Question : 67 of 120

Marks: +1, -0

What is the value of tan 18°?

Solution:

Concept:

Use complementary angle identities and trigonometric formulas to find

\tan 18^\circ

Explanation:

Let

\theta = 18^\circ

. Then

5\theta = 90^\circ

, so

2\theta = 90^\circ - 3\theta

Take sine on both sides:

\sin 2\theta = \sin(90^\circ - 3\theta) = \cos 3\theta

Apply formulas:

\sin 2\theta = 2\sin\theta\cos\theta

and

\cos 3\theta = 4\cos^3\theta - 3\cos\theta

Thus

2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta

Bring all terms:

2\sin\theta\cos\theta - 4\cos^3\theta + 3\cos\theta = 0

Factor

\cos\theta

\cos\theta(2\sin\theta - 4\cos^2\theta + 3) = 0

Since

\cos 18^\circ \neq 0

, we have

2\sin\theta - 4\cos^2\theta + 3 = 0

Replace

\cos^2\theta = 1 - \sin^2\theta

2\sin\theta - 4(1 - \sin^2\theta) + 3 = 0 \Rightarrow 4\sin^2\theta + 2\sin\theta - 1 = 0

Solve quadratic:

\sin\theta = \frac{-2 \pm \sqrt{4+16}}{8} = \frac{-1 \pm \sqrt{5}}{4}

Since

\theta

in first quadrant,

\sin 18^\circ > 0

, so

\sin\theta = \frac{\sqrt{5}-1}{4}

Now

\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{\sqrt{5}-1}{4}\right)^2} = \frac{\sqrt{10+2\sqrt{5}}}{4}

Therefore

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{(\sqrt{5}-1)/4}{(\sqrt{10+2\sqrt{5}})/4} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}

Answer:

Option A:

\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}

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