$\displaystyle sin^3(x+1)tan^4x + tan^2xsin^3(x+1)$ $\displaystyle = sin^3(x+1)tan^2x(tan^2x + 1)$ $\displaystyle = sin^3(x+1)tan^2x(1/(cos^2x))$ $\displaystyle = (sin^3(x+1)tan^2x)/(cos^2x)$
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Originally Posted by shane99 $\displaystyle sin^3(x+1)tan^4x + tan^2xsin^3(x+1)$ $\displaystyle = sin^3(x+1)tan^2x(tan^2x + 1)$ $\displaystyle = sin^3(x+1)tan^2x(1/(cos^2x))$ $\displaystyle = (sin^3(x+1)tan^2x)/(cos^2x)$ I see no problem with that. Remember that $\displaystyle \frac{1}{cos^2 \theta} = sec^2 \theta$
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