Question

solve this attachment?​

Answer 1

${tan}^{ - 1} ( \frac{acosx - bsinx}{bcosx + asinx} ) \\ \\ = {tan}^{ - 1} ( \frac{ \frac{acosx - bsinx}{bcosx} }{ \frac{bcosx + asinx}{bcosx} } ) \\ \\ = {tan}^{ - 1} ( \frac{ \frac{a}{b} - tanx }{1 + \frac{a}{b}tanx } ) \\ \\ = {tan}^{ - 1} \frac{a}{b} - {tan}^{ - 1} tanx \\ \\ = {tan}^{ - 1} \frac{a}{b} - x$