#### Question

Prove that:

\[\sec^4 A\left( 1 - \sin^4 A \right) - 2 \tan^2 A = 1\]

#### Solution

\[\sec^4 A\left( 1 - \sin^4 A \right) - 2 \tan^2 A\]

\[ = \sec^4 A - \sec^4 A \sin^4 A - 2 \tan^2 A\]

\[ = \left( 1 + \tan^2 A \right)^2 - \frac{\sin^4 A}{\cos^4 A} - 2 \tan^2 A \left( \sec^2 \theta = 1 + \tan^2 \theta\text{ and }\sec\theta = \frac{1}{\cos\theta} \right)\]

\[ = 1 + \tan^4 A + 2 \tan^2 A - \tan^4 A - 2 \tan^2 A \left[ \left( a + b \right)^2 = a^2 + b^2 + 2ab \right]\]

\[ = 1\]

Is there an error in this question or solution?

Solution Prave That: Sec 4 a ( 1 − Sin 4 a ) − 2 Tan 2 a = 1 Concept: Application of Trigonometry.