Question

Question #e1e9a

Answer 1

`"see explanation"`

Explanation:

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

`•color(white)(x)tanx=sinx/cosx`

`•color(white)(x)sin^2x+cos^2x=1`

`rArr(tan^2x)/(1+tan^2x)`

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

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

`=sin^2x/cos^2x xx(cos^2x)/1=sin^2xrArr"verified"`

Answer 2

We have: `frac(tan^(2)(theta))(1 + tan^(2)(theta))`

Let's apply the Pythagorean identity `1 + tan^(2)(theta) = sec^(2)(theta)`:

`= frac(tan^(2)(theta))(sec^(2)(theta))`

Then, let's apply two standard trigonometric identities; `tan(theta) = frac(sin(theta))(cos(theta))` and `sec(theta) = frac(1)(cos(theta))`:

`= frac(frac(sin^(2)(theta))(cos^(2)(theta)))(frac(1)(cos^(2)(theta)))`

`= frac(sin^(2)(theta))(cos^(2)(theta)) times frac(cos^(2)(theta))(1)`

`= frac(sin^(2)(theta) cos^(2)(theta))(cos^(2)(theta))`

Finally, let's cancel the `cos^(2)(theta)` terms:

`= sin^(2)(theta) " "" shown"`

Answer 3

See the explanation below

Explanation:

We need

`tanx=sinx/cosx`

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

Therefore,

`LHS=tan^2x/(1+tan^2x)`

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

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

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

`=sin^2x`

`=RHS`

`QED`

`**PART(b)**`

Let `tantheta=x`

`theta=tan^-1(x)`

`sin^2theta=x^2/(1+x^2)`

`sintheta=x/(sqrt(1+x^2))`

Therefore,

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

Drawing

Draw a right angle triangle

The hypotenuse is `=sqrt(1+x^2)`

The angle is `theta`

The opposite side is `=x`

The adjacent side is `=1`

Such that

`sintheta=x/sqrt(1+x^2)`

`tantheta=x/1=x`