If 7^{4}θ + 9^{4}θ + 42²θ = 16, 0

Correct Answer is : 13{1}{{3}}31

Correct Answer is : 13{1}{{3}}31

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UPSC CDS 2 2023 Math Paper

Question : 72 of 100

Marks: +1, -0

7\sin^{4}\theta + 9\cos^{4}\theta + 42\sin^{2}\theta = 16

0 < \theta < \frac{\pi}{2}

\tan \theta

equal to ?

Solution:

Concept:

Convert the trigonometric equation into a quadratic form in

\sin^2\theta

and solve for

\theta

Explanation:

Given

7\sin^4\theta + 9\cos^4\theta + 42\sin^2\theta = 16

, with

0 < \theta < \frac{\pi}{2}

Use

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

Then

\cos^4\theta = (1 - \sin^2\theta)^2 = 1 - 2\sin^2\theta + \sin^4\theta

Substitute into the equation:

7\sin^4\theta + 9(1 - 2\sin^2\theta + \sin^4\theta) + 42\sin^2\theta = 16

Simplify:

7\sin^4\theta + 9 - 18\sin^2\theta + 9\sin^4\theta + 42\sin^2\theta = 16

Combine like terms:

16\sin^4\theta + 24\sin^2\theta + 9 = 16

Thus

16\sin^4\theta + 24\sin^2\theta - 7 = 0

Let

x = \sin^2\theta

. Then

16x^2 + 24x - 7 = 0

Use the quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

with

a=16

b=24

c=-7

x = \frac{-24 \pm \sqrt{24^2 - 4 \cdot 16 \cdot (-7)}}{2 \cdot 16} = \frac{-24 \pm \sqrt{576 + 448}}{32} = \frac{-24 \pm \sqrt{1024}}{32} = \frac{-24 \pm 32}{32}

x = \frac{-24+32}{32} = \frac{8}{32} = \frac{1}{4}

x = \frac{-24-32}{32} = \frac{-56}{32} = -\frac{7}{4}

Since

0 < \theta < \frac{\pi}{2}

\sin\theta>0

, so

x = \sin^2\theta

must be positive. Hence

\sin^2\theta = \frac{1}{4}

Then

\sin\theta = \frac{1}{2}

, giving

\theta = 30^\circ

Therefore

\tan\theta = \tan 30^\circ = \frac{1}{\sqrt{3}}

Answer:

\boxed{\frac{1}{\sqrt{3}}}

(Option D).

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