#### Question

Prove that:

\[\sec^4 \theta - \cos^4 \theta = 1 - 2 \cos^2 \theta\]

#### Solution

**Disclaimer:** There is printing mistake in the question. The correct question should be

\[\sin^4 \theta - \cos^4 \theta = 1 - 2 \cos^2 \theta\].

The solution has been provided accordingly.

\[\sin^4 \theta - \cos^4 \theta\]

\[ = \left( \sin^2 \theta \right)^2 - \left( \cos^2 \theta \right)^2 \]

\[ = \left( \sin^2 \theta - \cos^2 \theta \right)\left( \sin^2 \theta + \cos^2 \theta \right) \left[ a^2 - b^2 = \left( a + b \right)\left( a - b \right) \right]\]

\[ = \sin^2 \theta - \cos^2 \theta \left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]

\[ = 1 - \cos^2 \theta - \cos^2 \theta\]

\[ = 1 - 2 \cos^2 \theta\]

The solution has been provided accordingly.

\[\sin^4 \theta - \cos^4 \theta\]

\[ = \left( \sin^2 \theta \right)^2 - \left( \cos^2 \theta \right)^2 \]

\[ = \left( \sin^2 \theta - \cos^2 \theta \right)\left( \sin^2 \theta + \cos^2 \theta \right) \left[ a^2 - b^2 = \left( a + b \right)\left( a - b \right) \right]\]

\[ = \sin^2 \theta - \cos^2 \theta \left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]

\[ = 1 - \cos^2 \theta - \cos^2 \theta\]

\[ = 1 - 2 \cos^2 \theta\]

Is there an error in this question or solution?

Solution Prave that: sec 4 θ − cos 4 θ = 1 − 2 cos 2 θ Concept: Application of Trigonometry.