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If sintheta +cos theta =m and sec theta ...

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

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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)`
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