#### Question

Prove the following.

sec^{6}x – tan^{6}x = 1 + 3sec^{2}x × tan^{2}x

#### Solution

We have,

\[\sec^2 x - \tan^2 x = 1\]

Cubing on both sides, we get

\[\left( \sec^2 x - \tan^2 x \right)^3 = 1^3 \]

\[ \Rightarrow \left( \sec^2 x \right)^3 - \left( \tan^2 x \right)^3 - 3 \times \sec^2 x \times \tan^2 x \times \left( \sec^2 x - \tan^2 x \right) = 1 \left[ \left( a - b \right)^3 = a^3 - b^3 - 3ab\left( a - b \right) \right]\]

\[ \Rightarrow \sec^6 x - \tan^6 x - 3 \sec^2 x \tan^2 x = 1\]

\[ \Rightarrow \sec^6 x - \tan^6 x = 1 + 3 \sec^2 x \tan^2 x\]

Is there an error in this question or solution?

Solution Prove the Following. Sec6x – Tan6x = 1 + 3sec2x × Tan2x Concept: Application of Trigonometry.