`tan^(-1)""[(a cos x -b sinx)/(b cosx+a sinx)]` को सरल कीजिए, यदि `""(a)/(b)tanx le-1`

Explanation

We have,
LHS `=tan^(-1)(acosx-bsinx)/(bcosx-asinx)=tan^(-1){((acosx-bsinx)/(bcosx))/((bcosx-asinx)/(bcosx))` [on dividing num. and denom. By `bcosx`].
`=tan^(-1){(a/b-tanx)/(1+a/btanx)}=tan^(-1)(p-q)/(1+pq)`, where `a/b=p` and `tanx=q`
`=tan^(-1)p-tan^(-1)q=tan^(-1)a/b-tan^(-1)(tanx)`
`=(tan^(-1)a/b-x)`= RHS.
Hence, `tan^(-1)((acosx-bsinx)/(bcosx-asinx))=(tan^(-1)a/b-x)`.