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Prove the Following. Sec6x – Tan6x = 1 + 3sec2x × Tan2x - Geometry

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Question

Prove the following.
sec6x – tan6x = 1 + 3sec2x × tan2x

Solution

We have,
\[\sec^2 x - \tan^2 x = 1\]
Cubing on both sides, we get
\[\left( \sec^2 x - \tan^2 x \right)^3 = 1^3 \]
\[ \Rightarrow \left( \sec^2 x \right)^3 - \left( \tan^2 x \right)^3 - 3 \times \sec^2 x \times \tan^2 x \times \left( \sec^2 x - \tan^2 x \right) = 1 \left[ \left( a - b \right)^3 = a^3 - b^3 - 3ab\left( a - b \right) \right]\]
\[ \Rightarrow \sec^6 x - \tan^6 x - 3 \sec^2 x \tan^2 x = 1\]
\[ \Rightarrow \sec^6 x - \tan^6 x = 1 + 3 \sec^2 x \tan^2 x\]

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APPEARS IN

 Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 6: Trigonometry
Problem set 6 | Q: 5.07 | Page no. 138
Solution Prove the Following. Sec6x – Tan6x = 1 + 3sec2x × Tan2x Concept: Application of Trigonometry.
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