Tan^-1[(acosx-bsinx)/(bcosx+asinx)]?

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Answer 1 · 2017-09-10T04:30:14

$tan^-1[(acosx-bsinx)/(bcosx+asinx)]$

$=tan^-1[((acosx)/(bcosx)-(bsinx)/(bcosx))/((bcosx)/(bcosx)+(asinx)/(bcosx))]$

$=tan^-1[(a/b-tanx)/(1+a/bxxtanx)]$

$=tan^-1[(tanalpha-tanx)/(1+tanalphaxxtanx)]$ ,

where $color(red)(a/b=tanalpha)$

$=tan^-1[tan(alpha-x)]$

$=alpha-x=tan^-1(a/b)-x$