`2+secxcscx=(sinx+cosx)^2/(sinxcosx)`

To prove, consider the left side of the equation.

`2+secxcscx`

Express the secant and cosecant in terms of cosine and sine, respectively.

`=2+1/cosx*1/sinx`

`=2+1/(sinxcosx)`

To add, express them as two fractions with same denominators.

`=2*(sinxcosx)/(sinxcosx)+1/(sinxcosx)`

`=(2sinxcosx)/(sinxcosx) + 1/(sinxcosx)`

`=(2sinxcosx + 1)/(sinxcosx)`

Apply the Pythagorean identity `sin^2x+cos^2x=1` .

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

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

And, factor the numerator.

`= ((sinx +cosx)(sinx+cosx))/(sinxcosx)`

`=(sinx+cosx)^2/(sinxcosx)`

Notice that this is the same expression that the right side of the equation have. **Thus, this proves that the `2+secxcscx=(sinx+cosx)^2/(sinxcosx)` is an identity.**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now