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NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If ^ 6 θ + ^ 6 θ + k ^ 2 2θ = 1, then k is equal to

If

\sin^{6} \theta + \cos^{6} \theta + k\cos^{2} 2\theta = 1,

then

k

is equal to

Options

A $\dfrac{1}{2}\tan^{2} 2\theta$
B $\dfrac{1}{4}\tan^{2} 2\theta$
C $4\cot^{2} 2\theta$
D $\dfrac{3}{4}\tan^{2} 2\theta$

Correct answer

D. $\dfrac{3}{4}\tan^{2} 2\theta$

Step-by-step solution

\left(\sin^{2} \theta + \cos^{2} \theta\right)\left(\sin^{4} \theta - \sin^{2} \theta{cos}^{2} \theta + \cos^{4} \theta\right) + k{cos}^{2} 2\theta = 1

or

{\left(\sin^{2} \theta + \cos^{2} \theta\right)}^{2} - 3\sin^{2} \theta{cos}^{2} \theta + k{cos}^{2} 2\theta = 1

or

1 - 3\sin^{2} \theta{cos}^{2} \theta + k{cos}^{2} 2\theta = 1

or

k{cos}^{2} 2\theta = 3\sin^{2} \theta{cos}^{2} \theta

or

k\left(\cos^{2} 2\theta\right) = \dfrac{3}{4}{\left(2sin \theta cos \theta\right)}^{2}

or

k{cos}^{2} 2\theta = \dfrac{3}{4}\sin^{2} 2\theta

or

k = \dfrac{3}{4}\tan^{2} 2\theta

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