Prove that:sec⁶A - tan⁶A = 1 + 3 tan²A + 3 tan⁴A

1.	Prove that:sec⁶A - tan⁶A = 1 + 3 tan²A + 3 tan⁴A
Answer» $\texttt{\textsf{\large{\underline{Solution}:}}}$ We have to prove: → sec⁶A - tan⁶A = 1 + 3 tan²A + 3 tan⁴A Taking LHS, we get: = sec⁶A - tan⁶A = (sec²A)³ - (tan²A)³ Using identity a³ - b³ = (a - b)³ + 3ab(a - b), we get: = (sec²A - tan²A)³ + 3 × sec²A × tan²A(sec²A - tan²A) We know that: → sec²θ - tan²θ = 1 Therefore, we get: = 1³ + 3 × sec²A × tan²A = 1 + 3 × (1 + tan²A) × tan²A [sec²θ = 1 + tan²θ] = 1 + 3 × (tan²A + tan⁴A) = 1 + 3 tan²A + 3 tan⁴A = RHS (HENCE Proved) $\texttt{\textsf{\large{\underline{Know More}:}}}$ 1. Relationship between SIDES. sin(x) = Height/Hypotenuse. COS(x) = Base/Hypotenuse. tan(x) = Height/Base. cot(x) = Base/Height. sec(x) = Hypotenuse/Base. cosec(x) = Hypotenuse/Height. 2. Square FORMULAE. sin²x + cos²x = 1. cosec²x - cot²x = 1. sec²x - tan²x = 1 3. Reciprocal Relationship. sin(x) = 1/cosec(x). cos(x) = 1/sec(x). tan(x) = 1/cot(x). 4. Cofunction identities. sin(90° - x) = cos(x) and vice versa. cosec(90° - x) = sec(x) and vice versa. tan(90° - x) = cot(x) and vice versa.

Answer»

$\texttt{\textsf{\large{\underline{Solution}:}}}$

We have to prove:

→ sec⁶A - tan⁶A = 1 + 3 tan²A + 3 tan⁴A

Taking LHS, we get:

= sec⁶A - tan⁶A

= (sec²A)³ - (tan²A)³

Using identity a³ - b³ = (a - b)³ + 3ab(a - b), we get:

= (sec²A - tan²A)³ + 3 × sec²A × tan²A(sec²A - tan²A)

We know that:

→ sec²θ - tan²θ = 1

Therefore, we get:

= 1³ + 3 × sec²A × tan²A

= 1 + 3 × (1 + tan²A) × tan²A [sec²θ = 1 + tan²θ]

= 1 + 3 × (tan²A + tan⁴A)

= 1 + 3 tan²A + 3 tan⁴A

= RHS (HENCE Proved)

$\texttt{\textsf{\large{\underline{Know More}:}}}$

1. Relationship between SIDES.

2. Square FORMULAE.

3. Reciprocal Relationship.

4. Cofunction identities.

Discussion