sec^6A-tan^2A=1+3 tan^2A+ 3 tan^4A
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Hi ,
We know that the trigonometric
identity
Sec² A = 1 + tan² A
LHS = sec ^6 A - tan² A
= ( Sec² A )³ - tan² A
= ( 1 +tan² A)³ - tan² A
= 1³ + 3×(tan²A)²×1+3×tan²A×1²+
(tan² A )³ - tan² A
= 1 + 3tan^4A + 3tan²A + tan^6 A
- tan² A
= tan^6 A + 3tan^4 A + 2tan² A + 1
= RHS
I hope this helps you.
:)
We know that the trigonometric
identity
Sec² A = 1 + tan² A
LHS = sec ^6 A - tan² A
= ( Sec² A )³ - tan² A
= ( 1 +tan² A)³ - tan² A
= 1³ + 3×(tan²A)²×1+3×tan²A×1²+
(tan² A )³ - tan² A
= 1 + 3tan^4A + 3tan²A + tan^6 A
- tan² A
= tan^6 A + 3tan^4 A + 2tan² A + 1
= RHS
I hope this helps you.
:)
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