Question

Question #723dc

Answer 1

Please see below.

Explanation:

.

We have the identity:

`sec^2x=tan^2x+1`

Let's plug this into the numerator and factor the `tanx` out of the denominator:

`(tan^2x+1-1)/(tanx(sin^2x+cos^2x))=tan^2x/(tanx(sin^2x+cos^2x))=`

`tanx/(sin^2x+cos^2x)`

But we know:

`sin^2x+cos^2x=1`

Therefore,

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

Answer 2

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)tan^2x+1=sec^2xrArrtan^2x=sec^2x-1`

`•color(white)(x)sin^2x+cos^2x=1`

`"consider the left hand side"`

`((sec^2x-1))/(tanxsin^2x+tanxcos^2x)`

`=(tan^2x)/(tanx(sin^2x+cos^2x))`

`=tan^2x/tanx`

`=tanx=" right hand side "rArr"verified"`