If `tan theta=1`, then `(sin theta+cos theta)/(sec theta+cosec theta)=`

Explanation

`tan theta=1`
From the table of trigonometric ratios,
We know that `tan 45^(@)=1`
`:.theta=45^(@)`
`sin theta=sin 45^(@)=1/(sqrt(2))`
`cos theta=cos45^(@)=1/(sqrt(2))`
`sec theta =sec 45^(@)=sqrt(2)`
`cosec theta=cosec 45^(@)=sqrt(2)`
`(sin theta+cos theta)/(sec theta+cosec theta)=(sin 45^(@)+cos 45^(@))/(sec 45^(@)+cosec 45^(@))`
`=(1/(sqrt(2))+1/(sqrt(2)))/(sqrt(2)+sqrt(2))`
`=((2/(sqrt(2))))/(sqrt(2))`
`=2/(sqrt(2))-:2sqrt(2)`
`=2/(sqrt(2))xx1/(2sqrt(2))`
`=1/2`
`(sin theta+cos theta)/(sec theta+cosec theta)=1/2`