# Show that tan(θ) + cot(θ) = sec(θ)csc(θ) You also may use the alternative approach, such that:

`cot theta = 1/tan theta`

Replacing `1/tan theta` for `cot theta` yields:

`tan theta + 1/tan theta = sec theta csc theta`

Bringing the terms to the left side to a common denominator, yields:

`(tan^2 theta + 1)/tan theta = sec...

You also may use the alternative approach, such that:

`cot theta = 1/tan theta`

Replacing `1/tan theta` for `cot theta` yields:

`tan theta + 1/tan theta = sec theta csc theta`

Bringing the terms to the left side to a common denominator, yields:

`(tan^2 theta + 1)/tan theta = sec theta csc theta`

Using the trigonometric identity `tan^2 theta + 1 = sec^2 theta` yields:

`(sec^2 theta)/tan theta = sec theta csc theta`

Reducing duplicate factors both sides yields:

`sec theta/tan theta = csc theta`

Replacing `1/cos theta` for `sec theta ` yields:

`(1/cos theta) = tan theta*csc theta`

Replacing `sin theta/cos theta` for `tan theta` and `1/sin theta` for `csc theta` , yields:

`(1/cos theta) = (sin theta/cos theta)*(1/sin theta)`

Reducing duplicate factors yields:

`(1/cos theta) = 1/cos theta`

Hence, evaluating both sides yields the same results, thus, the given identity `tan theta + cot theta = sec theta*csc theta` holds.

Approved by eNotes Editorial Team The trigonometric identity `tan theta + cot theta = sec theta*cosec theta` has to be proved.

Start from the left hand side:

`tan theta + cot theta`

= `(sin theta)/(cos theta) + (cos theta)/(sin theta)`

= `((sin theta)*(sin theta) + (cos theta)(cos theta))/(cos theta*sin theta)`

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

= `1/(cos theta*sin theta)`

= `(1/(cos theta))*(1/(sin theta))`

= `sec theta*cosec theta`

This proves the identity: `tan theta + cot theta = sec theta*cosec theta`

Approved by eNotes Editorial Team