If `sintheta +cos theta =m` and `sec theta + cosec theta =n` then `n(m+1)(m-1)` equals

Explanation

Here, `sintheta+costheta = m`

`sectheta+cosectheta = n`

`:. n(m+1)(m-1) = n(m^2-1)`

`= (sectheta+cosectheta)((sintheta+costheta)^2-1)`

`= (1/costheta+1/sintheta)(sin^2theta+cos^2theta+2sinthetacostheta-1)`

`= ((sin theta+costheta)/(sinthetacostheta))(1+2sinthetacostheta-1)`

`=(2(sintheta+costheta)sinthetacostheta)/(sinthetacostheta)`

`=(2(sintheta+costheta)`

...