Question

sec⁴x tan⁶x=(tan⁶x + tan⁸x)sec²x
note that each statement must be based on a rule chosen from the rule menu. to see a det the right of the rule.
statement reason
=(1 + tan²x)²tan⁶x pythagorean
=(1 + 2tan²x + tan⁴x)tan⁶x algebra
= select rule
validate

Explanation:

Step1: Expand the product

$(1 + 2\tan^{2}x+\tan^{4}x)\tan^{6}x=1\times\tan^{6}x + 2\tan^{2}x\times\tan^{6}x+\tan^{4}x\times\tan^{6}x$

Step2: Use exponent - rule for same - base powers ($a^m\times a^n=a^{m + n}$)

$1\times\tan^{6}x + 2\tan^{2}x\times\tan^{6}x+\tan^{4}x\times\tan^{6}x=\tan^{6}x+2\tan^{8}x+\tan^{10}x$

Step3: Factor out $\sec^{2}x$

We know that $\sec^{2}x = 1+\tan^{2}x$. And $\tan^{6}x+\tan^{8}x=\tan^{6}x(1 + \tan^{2}x)=\tan^{6}x\sec^{2}x$.
The original left - hand side $\sec^{4}x\tan^{6}x=(1 + \tan^{2}x)^{2}\tan^{6}x=(1 + 2\tan^{2}x+\tan^{4}x)\tan^{6}x=\tan^{6}x+2\tan^{8}x+\tan^{10}x$.
The right - hand side $(\tan^{6}x+\tan^{8}x)\sec^{2}x=\tan^{6}x\sec^{2}x+\tan^{8}x\sec^{2}x$.
If we continue from $(1 + 2\tan^{2}x+\tan^{4}x)\tan^{6}x$:
\[

$$\begin{align*} (1 + 2\tan^{2}x+\tan^{4}x)\tan^{6}x&=\tan^{6}x+2\tan^{8}x+\tan^{10}x\\ &=(\tan^{6}x+\tan^{8}x)+ \tan^{8}x+\tan^{10}x\\ &=(\tan^{6}x+\tan^{8}x)(1 + \tan^{2}x)\\ &=(\tan^{6}x+\tan^{8}x)\sec^{2}x \end{align*}$$

\]

Answer:

$\tan^{6}x + 2\tan^{8}x+\tan^{10}x$