Question

How do you verify the identify `sinthetatantheta+costheta=sectheta`?

Answer 1

See the proof below

Explanation:

We need

`tantheta=sintheta/costheta`

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

`sectheta=1/costheta`

Therefore,

`LHS=sinthetatantheta+costheta`

`=sintheta*sintheta/costheta+costheta`

`=(sin^2theta+cos^2theta)/(costheta)`

`=1/costheta`

`=sectheta`

`=RHS`

`QED`

Answer 2

Apply the identities `tan(theta)=(sin theta)/(cos theta)` and `sec theta=1/(cos theta)` along with the Pythagorean theorem.

Explanation:

Apply the identity `tan(theta)=(sin theta)/(cos theta)`:
`L.H.S.=sin theta*sin theta/(cos theta)+cos theta`
`=sin^2 theta/(cos theta)+cos theta`
`=(sin^2 theta+cos^2 theta)/(cos theta)`

Apply the Pythagorean theorem `sin^2 theta+cos^2 theta=1`
`=(sin^2 theta+cos^2 theta)/(cos theta)`
`=1/(cos theta)`

By the definition of secants `1/(cos theta)=sec theta`:
`=sec theta`
`=R.H.S`

Answer 3

`"see explanation"`

Explanation:

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

`•color(white)(x)tantheta=sintheta/costheta" and "sectheta=1/costheta"`

`•color(white)(x)sin^2theta+cos^2theta=1`

`"Consider the left side"`

`rArrsinthetatantheta+costheta`

`=sinthetaxxsintheta/costheta+costheta`

`=sin^2theta/costheta+cos^2theta/costhetalarr"common denominator "costheta`

`=(sin^2theta+cos^2theta)/costheta`

`=1/costheta=sectheta=" right side "rArr" verified"`