Answer:
Given, tan \[(A+B)=\sqrt{3},\tan (A-B)=\frac{1}{\sqrt{3}}\] \[\Rightarrow \tan (A+B)=\tan 60{}^\circ \] \[(A+B)=60{}^\circ \] ?(i) And, \[\tan (A-B)=\tan 30{}^\circ \] \[(A-B)=30{}^\circ \] ?(ii) On adding eq. (i) & (ii) \[_{\begin{smallmatrix} A\,-\,B\,=\,30{}^\circ \\ \overline{\underline{2\,A\,\,\,\,\,\,=\,90{}^\circ \,}} \end{smallmatrix}}^{A\,+\,B\,=\,60{}^\circ }\] [By adding] \[\Rightarrow A=\frac{90{}^\circ }{2}=45{}^\circ \] From. eq. (i), \[A+B=60{}^\circ \] \[45{}^\circ +B=60{}^\circ \] \[B=15{}^\circ \] \[\therefore A~=45{}^\circ ,\,B=15{}^\circ \] Now, \[tan\text{ }A.\text{ }sin\text{ (}A+B)+cos\text{ }A.\text{ }tan\text{ (}A-B)\] \[=\tan \text{ }45{}^\circ .\text{ }sin\text{ (}60{}^\circ )+cos\text{ }45{}^\circ ,\text{ }tan\text{ (}30{}^\circ )\] \[=1\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{3}}\] \[=\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{3}}{2}+\frac{\sqrt{6}}{6}\] \[=\frac{3\sqrt{3}+\sqrt{6}}{6}\]
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