Question

$1+4x^2 = 4x.secA$

Sec⁶A - tan⁶A =?

Answer 1

Given,

$1+4x^2=4xsecA$ , so write $secA$ in terms of $x$ as $secA=\frac{1+4x^2}{4x}$

Use trigonometry to find $p$ in terms of $x$ as,

$p=\sqrt{(1+4x^2)^2-(4x)^2}$

$=\sqrt{1+16x^4+8x^2-16x^2}$

$=\sqrt{1+16x^4-8x^2}$

$=\sqrt{1^2+(4x^2)^2-2(1)(4x^2)}$

$=\sqrt{(1-4x^2)^2}$

$=1-4x^2$

So, $tanA=\frac{1-4x^2}{4x}$

Now, simplify the expression $sec^6A-tan^6A$ as,

$sec^6A-tan^6A=(sec^2A)^3-(tan^2A)^3$

Use formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ to obtain,

$sec^6A-tan^6A=(sec^2A-tan^2A)(sec^4A+sec^2Atan^2A+tan^4A)$

$=(1)(sec^4A+sec^2Atan^2A+tan^4A)$

$=sec^4A+sec^2Atan^2A+tan^4A$

$=sec^4A+sec^2Atan^2A+(sec^2A-1)^2$

$=sec^4A+sec^2Atan^2A+sec^4A-2sec^2A+1$

$=2sec^4A+sec^2Atan^2A-2sec^2A+1$

$=2sec^2A(sec^2A-1)+sec^2Atan^2A+1$

$=2sec^2Atan^2A+sec^2Atan^2A+1$

$=3sec^2Atan^2A+1$

Substitute $secA=\frac{1+4x^2}{4x}$ and $tanA=\frac{1-4x^2}{4x}$ into simplified expression to obtain,

$=3(\frac{1+4x^2}{4x})^2(\frac{1-4x^2}{4x})^2+1$

$=3[(\frac{1+4x^2}{4x})(\frac{1-4x^2}{4x})]^2+1$

$=3[\frac{(1+4x^2)(1-4x^2)}{16x^2}]^2+1$

$=3[\frac{1-16x^4}{16x^2}]^2+1$

$=3(\frac{1+256x^8-32x^4}{256x^4})+1$

$=\frac{3+768x^8-96x^4+256x^4}{256x^4}$

$=\frac{768x^8+160x^4+3}{256x^4}$

Therefore, the simplified expression in terms of $x$ is $sec^6A-tan^6A==\frac{768x^8+160x^4+3}{256x^4}$