Question

sec⁶A - tan⁶A = 1+3sec²A.tan²A​

Answer 1

Answer:

hope it is helpful ...........

Answer 2

To prove :

• sec⁶A - tan⁶A = 1 + 3 sec⁶A . tan²A

Proof :

Taking LHS

→ LHS = sec⁶A - tan⁶A

→ LHS = (sec²A)³ - (tan²A)³

using trigonometric identity

1 + tan²A = sec²A

→ LHS = ( 1 + tan²A )³ - (tan²A)³

using algebraic identity

( a + b )³ = a³ + b³ + 3 ab ( a + b )

→ LHS = [ (1)³ + (tan²A)³ + 3 (1) (tan²A) ( 1 + tan²A) ] - (tan²A)³

→ LHS = 1 + tan⁶A + 3 tan²A ( 1 + tan²A ) - tan⁶A

→ LHS = 1 + 3 tan²A ( 1 + tan²A )

using trigonometric identity

1 + tan²A = sec²A

→ LHS = 1 + 3 tan²A sec²A = RHS

Proved .

More trigonometric identities and ratios

• sin²Ф + cos²Ф = 1
• 1 + tan²Ф = sec²Ф
• 1 + cot²Ф = cosec²Ф

• sin (90 - Ф) = cos Ф
• cos (90 - Ф) = sin Ф
• tan (90 - Ф) = cot Ф
• cot (90 - Ф) = tan Ф
• cosec (90 - Ф) = sec Ф
• sec (90 - Ф) = cosec Ф