question_answer

If tan\[(A+B)=\sqrt{3}\] and tan\[(A-B)=\frac{1}{\sqrt{3}}\] , where\[0<A+B<90{}^\circ ,\text{ }A>B\], find A and B. Also calculate\[\tan \text{ }A.\text{ }\sin \text{ }(A+B)+\cos \text{ }A.\text{ }\tan \text{ }(A-B)\].

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, \[tan\text{ }A.\text{ }sin\text{ (}A+B)+cos\text{ }A.\text{ }tan\text{ (}A-B)\]

\[=\tan \text{ }45{}^\circ .\text{ }sin\text{ (}60{}^\circ )+cos\text{ }45{}^\circ ,\text{ }tan\text{ (}30{}^\circ )\]

\[=1\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{3}}\]

\[=\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{3}}{2}+\frac{\sqrt{6}}{6}\]

\[=\frac{3\sqrt{3}+\sqrt{6}}{6}\]