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If s inalphas inbeta-cosalphacosbeta+1=0...

If `s inalphas inbeta-cosalphacosbeta+1=0,` then prove that `1+cotalphatanbeta=0`

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Given,
`sinalphasinbeta−cosalphacosbeta+1=0`
=`cosalphacosbeta−sinalphasinbeta=1 `
=` cos(alpha+beta)=1`
so, `(alpha+beta)=0`
now, ` tan(alpha+beta)=0 `
...
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