Question

How do you simplify `(tanx + 1)^2 cosx`?

Answer 1

`secx(1+sin2x)`

Explanation:

Let's begin by expanding the bracket.

`(tanx+1)^2=tan^2x+2tanx+1`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(tanx=(sinx)/(cosx))color(white)(a/a)|)))`

`rArrtan^2x+2tanx+1=(sin^2x)/(cos^2x)+(2sinx)/cosx+1`

substitute this back into the original.

That is `(sin^2x/cos^2x+(2sinx)/cosx+1)cosx`

distribute the bracket.

`rArrsin^2x/cosx+2sinx+cosx`

express each term with a common denominator of cosx

`=sin^2x/cosx+(2sinxcosx)/cosx+cos^2x/cosx`

`=(sin^2x+2sinxcosx+cos^2x)/(cosx)........ (A)`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(sin^2x+cos^2x=1)|)))`

and `color(red)(bar(ul(|color(white)(a/a)color(black)(sin2x=2sinxcosx)color(white)(a/a)|)))`

Returning to (A)

`rArr(1+sin2x)/cosx=1/cosx+(sin2x)/cosx`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(secx=1/(cosx))color(white)(a/a)|)))`

`rArr1/cosx+(sin2x)/cosx=secx+secxsin2x`

`rArr(tanx+1)^2cosx=secx(1+sin2x)`