Prove the following. sec^6 x – tan^6 x = 1 + 3 sec^2 x * tan^2 x.
Chapter 6 – Trigonometry – Text Book Solution
Problem set 6| Q 5.7| Page 138
Prove the following. sec^6 x – tan^6 x = 1 + 3 sec^2 x * tan^2 x.
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 – 3ableft( 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]
Solution
We have, sec^2 x – tan^2 x = 1 Cubing on both sides, we get (sec^2 x – tan^2 x)^3 = 1^3 => (sec^2 x)^3 – (tan^2 x)^3 – 3 * sec^2 x * tan^2 x * (sec^2 x – tan^2 x) = 1 [ (a – b)^3 = a^3 – b^3 – 3ab(a – b) ] => sec^6 x – tan^6 x – 3 sec^2 x tan^2 x = 1 => sec^6 x – tan^6 x = 1 + 3 sec^2 x tan^2 x
Chapter 6 – Trigonometry – Text Book Solution
Problem Set 6 |Q 5.7| P 138
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