Question

Question #45dfa

Answer 1

The 3 first expressions are lacking some parentheses. I'm not sure if I put the missing parentheses in the right places, but in my answer the identities hold true.

Explanation:

1) `(1-cos^2 theta)/(sec^2 theta-tan^2 theta+costheta)+costheta=1`

`(1-cos^2 theta)/(1/cos^2theta-sin^2theta/cos^2theta+costheta)+costheta=1`
`(1-cos^2theta)/((1-sin^2theta)/cos^2theta+costheta)+costheta=1`
`(1-cos^2theta)/(cancel(cos^2theta)/cancel(cos^2theta)+costheta)+costheta=1`
`((1-costheta)cancel((1+costheta)))/cancel(1+costheta)+costheta=1`
`1-cancel(costheta)+cancel(costheta)=1`
`1=1` (That's true and therefore the original identity holds true)

2) `(sin^2x+4sinx+3)/cos^2x=(sinx+3)/(1-sinx)`

`(sin^2x+4sinx+3)/cancel(cos^2x)=((sinx+3)(1+sinx))/cancel((1-sinx)(1+sinx))`
`sin^2x+4sinx+3=sinx+sin^2x+3+3sinx`
`sin^2x+4sinx+3=sin^2x+4sinx+3` (That's true and therefore the original identity holds true)

3) `(sectheta*sintheta)/(tantheta+cottheta)=sin^2theta`

`1/costheta*sintheta/(sintheta/costheta+costheta/sintheta)=sin^2theta`
`1/cancel(costheta)*sintheta/((sin^2theta+cos^2theta)/(cancel(costheta)sintheta))=sin^2theta`
`sintheta*sintheta=sin^2theta` (That's true and therefore the original identity holds true)

4) `csc^2theta*tan^2theta-1=tan^2theta`

`1/cancel(sin^2theta)*cancel(sin^2theta)/cos^2theta-1=tan^2theta`
`sec^2theta-1=tan^2theta` (We know that this is true but we can go on)
`1/cos^2theta-1=sin^2theta/cos^2theta`
`(1-cos^2theta)/cancel(cos^2theta)=sin^2theta/cancel(cos^2theta)`
`1=sin^2theta+cos^2theta` (That's true and therefore the original identity holds true)