Question

Sec⁶A+tan⁶A=(2sec²A−1)(sec⁴A−sec²A+1)​

Answer 1

See Below.

Explanation:

We want to verify that:

`\displaystyle \sec^6 A + \tan ^6 A = (2 \sec^2 A - 1)( \sec ^4 A - \sec^2 A + 1)`

Recall that:

`\displaystyle (a^3 + b^3) = (a+b)(a^2 - ab + b^2)`

Let a = sec²(A) and b = tan²(A). This yields:

`\displaystyle \sec^6 A + \tan ^6 A = (\sec ^2 A + \tan^2 A)(\sec^4 A - \sec^2 A \tan^2 A + \tan^4 A)`

Recall that from the Pythagorean Identity, tan²(θ) + 1 = sec²(θ). Hence:

`\displaystyle \begin{aligned}\displaystyle \sec^6 A + \tan ^6 A &= (\sec ^2 A + \tan^2 A)(\sec^4 A - \sec^2 A \tan^2 A + \tan^4 A) \\ \\ &=(\sec ^2 A + (\sec^2 A -1))(\sec^4 A - \sec^2 A \tan^2 A + \tan^4 A) \\ \\ &= (2\sec^2 A - 1)(\sec^4 A - \sec^2 A \tan^2 A + \tan^4 A) \end{aligned}`

Likewise:

`\displaystyle \begin{aligned}\sec^4 A - & \sec^2 A \tan^2 A + \tan^4 A = \sec^4 A - \sec^2 A (\sec^2 A - 1)+ (\sec^2 A -1)^2\\ \\ &=\sec^4 A - \sec^4 A + \sec^2 A + (\sec^4 A - 2\sec^2 A + 1) \\ \\ &= \sec^4 A -\sec^2 A + 1\end{aligned}`

Hence:

`\displaystyle \begin{aligned}\displaystyle \sec^6 A + \tan ^6 A &= (2\sec^2 A - 1)(\sec^4 A - \sec^2 A \tan^2 A + \tan^4 A) \\ \\ &= (2\sec^2 A -1)(\sec^4 A - \sec^2 A + 1)\\ \\ &\stackrel{\checkmark}{=} (2\sec^2 A -1)(\sec^4 A - \sec^2 A + 1)\end{aligned}`

Hence verified.