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

Question

40779 views around the world

You can reuse this answer
Creative Commons License

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#