Prove that : Sec^6x-tan^6x= 1+3tan^2x+ 3tan^4x
Asked by siddharthlathia0
10th June 2018,
6:55 AM
Answered by Expert
Answer:
Check your question
a³-b³=(a-b)³+3ab(a-b)
1+tan²x=sec²x
(sec²x)³-(tan²x)³=(sec²x-tan²x)³+3sec²xtan²x(sec²x-tan²x)
=1³+3sec²xtan2x(1)
=1+3tan²xsec²x
1+tan²x=sec²x
(sec²x)³-(tan²x)³=(sec²x-tan²x)³+3sec²xtan²x(sec²x-tan²x)
=1³+3sec²xtan2x(1)
=1+3tan²xsec²x
Answered by Expert
11th June 2018,
11:07 AM
Rate this answer
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
You have rated this answer /10
Your answer has been posted successfully!
RELATED STUDY RESOURCES :
Browse free questions and answers by Chapters
- 1 Sets
- 2 Relations and Functions
- 3 Principle of Mathematical Induction
- 4 Binomial Theorem
- 5 Sequences and Series
- 6 Straight Lines
- 7 Conic Sections
- 8 Introduction to Three Dimensional Geometry
- 9 Limits and Derivatives
- 10 Mathematical Reasoning
- 11 Statistics
- 12 Probability
- 13 Trigonometric Functions
- 14 Complex Numbers and Quadratic Equations
- 15 Permutations and Combinations
- 16 Linear Inequalities
Trending Tags
Latest Questions
JEE Main Mathematics
Asked by amaladelsingh
28th October 2020,
10:24 PM
CBSE X Biology
Asked by reenacharan2008
28th October 2020,
9:11 PM
