If ^{8} θ + ^{8} θ = m, then what is the value of θ + θ ?

Correct Answer is : m+2+2+2{{{m+2}+2}+2}m+2+2+2

Correct Answer is : m+2+2+2{{{m+2}+2}+2}m+2+2+2

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UPSC CDS 1 2023 Math Paper

Question : 48 of 100

Marks: +1, -0

\tan^{8} \theta + \cot^{8} \theta = m

\tan \theta + \cot \theta

Solution:

Concept:

Repeatedly apply the identity

\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2

to reduce powers from

8

1

Explanation:

Let

t = \tan \theta

. Then

\cot \theta = \frac{1}{t}

Given:

t^8 + \frac{1}{t^8} = m

First, consider

t^4 + \frac{1}{t^4}

\left(t^4 + \frac{1}{t^4}\right)^2 = t^8 + \frac{1}{t^8} + 2 = m + 2

t^4 + \frac{1}{t^4} = \sqrt{m+2}

Next, consider

t^2 + \frac{1}{t^2}

\left(t^2 + \frac{1}{t^2}\right)^2 = t^4 + \frac{1}{t^4} + 2 = \sqrt{m+2} + 2

t^2 + \frac{1}{t^2} = \sqrt{\sqrt{m+2} + 2}

Finally, consider

t + \frac{1}{t}

\left(t + \frac{1}{t}\right)^2 = t^2 + \frac{1}{t^2} + 2 = \sqrt{\sqrt{m+2} + 2} + 2

t + \frac{1}{t} = \sqrt{\sqrt{\sqrt{m+2} + 2} + 2}

Since

t = \tan \theta

, we have

\tan \theta + \cot \theta = \sqrt{\sqrt{\sqrt{m+2} + 2} + 2}

Answer:

\sqrt{\sqrt{\sqrt{m+2} + 2} + 2}

(Option C).

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