{vmatrix} ²θ & -²θ \\ ²θ & ²θ {vmatrix} =

Correct Answer is : 14(3+cos⁡4θ)1/4(3 + 4θ)41(3+cos4θ)

Correct Answer is : 14(3+cos⁡4θ)1/4(3 + 4θ)41(3+cos4θ)

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GUJCET 2025 Math Paper

Question : 15 of 40

Marks: +1, -0

\begin{vmatrix} \cos^2\theta & -\sin^2\theta \\ \sin^2\theta & \cos^2\theta \end{vmatrix} = \dots

Solution:

Concept:

Use determinant expansion and trigonometric identities to simplify the expression.

Explanation:

Calculate the determinant:

\begin{vmatrix}\cos^2\theta & -\sin^2\theta\\ \sin^2\theta & \cos^2\theta\end{vmatrix}=\cos^2\theta\cos^2\theta-(-\sin^2\theta)\sin^2\theta

=\cos^4\theta+\sin^4\theta

Use

a^2+b^2=(a+b)^2-2ab

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

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

Since

\sin2\theta=2\sin\theta\cos\theta

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

2\sin^2\theta\cos^2\theta=\frac12\sin^2 2\theta

Thus, determinant

=1-\frac12\sin^2 2\theta

Now use

\sin^2 2\theta=\frac{1-\cos4\theta}{2}

1-\frac12\left(\frac{1-\cos4\theta}{2}\right)=1-\frac14+\frac14\cos4\theta

=\frac34+\frac14\cos4\theta=\frac14(3+\cos4\theta)

So option B is correct.

Answer:

\frac14(3+\cos4\theta)

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