since, \(\alpha \),\(\beta \) lie between 0 and \(\frac { \pi }{ 4 } \)
\(\therefore -\frac { \pi }{ 4 } <\alpha -\beta \frac { \pi }{ 4 } and\quad 0<\alpha +\beta <\frac { \pi }{ 2 } \)
\(\Rightarrow cos\quad (\alpha -\beta )\quad and\quad sin\quad (\alpha +\beta )\quad are\quad positive.\)
\(Now,\quad sin\quad (\alpha +\beta )=\sqrt { 1-{ cos }^{ 2 }\left( \alpha +\beta \right) } =\frac { 3 }{ 5 } \)
\(and\quad \quad cos\quad (\alpha -\beta )=\sqrt { 1-{ sin }^{ 2 }\left( \alpha -\beta \right) } =\frac { 12 }{ 13 } \)
\(\therefore \quad tan\quad (\alpha +\beta )=\frac { sin\left( \alpha +\beta \right) }{ cos\left( \alpha +\beta \right) } =\frac { 3/5 }{ 4/5 } =\frac { 3 }{ 4 } \)
\(and\quad tan\quad (\alpha -\beta )=\frac { sin\quad \left( \alpha -\beta \right) }{ cos\quad \left( \alpha -\beta \right) } =\frac { 5/13 }{ 12/13 } =\frac { 5 }{ 12 }\)
\(now,\quad tan\quad 2\alpha =tan\quad [\left( \alpha +\beta \right) +\left( \alpha -\beta \right) ]\)
\(=\frac { tan\quad (\alpha +\beta )+tan\quad (\alpha -\beta ) }{ 1-tan\quad (\alpha +\beta )tan\left( \alpha -\beta \right) } =\frac { \frac { 3 }{ 4 } +\frac { 5 }{ 12 } }{ 1-\frac { 3 }{ 4 } \times \frac { 5 }{ 12 } } =\frac { 56 }{ 33 } \)
Hence proved.

\(Given,\quad tan\quad \theta =k\quad tan\quad \phi \)
\(\Rightarrow \frac { tan\quad \theta }{ tan\quad \phi } =\frac { k }{ 1 } =\frac { tan\quad \theta +tan\quad \phi }{ tan\quad \theta -tan\quad \phi } =\frac { k+1 }{ k-1 } \) [applying componendo and dividendo]
\(\Rightarrow \frac { \frac { sin\quad \theta }{ cos\quad \theta } +\frac { sin\quad \phi }{ cos\quad \phi } }{ \frac { sin\quad \theta }{ cos\quad \theta } -\frac { sin\quad \phi }{ cos\quad \phi } } =\frac { k+1 }{ k-1 }\)
\(\Rightarrow \frac { sin\quad \theta .cos\quad \phi +cos\quad \theta .sin\quad \phi }{ sin\quad \theta .cos\quad \phi -cos\quad \theta .sin\quad \phi } =\frac { k+1 }{ k-1 }\)
\(\Rightarrow \frac { sin\quad \left( \theta +\phi \right) }{ sin\quad \left( \theta -\phi \right) } =\frac { k+1 }{ k-1 } \Rightarrow \frac { sin\quad \left( \theta -\phi \right) }{ sin\quad \left( \theta +\phi \right) } =\frac { k-1 }{ k+1 } \)
\(\Rightarrow sin\quad (\theta -\phi )=\frac { k-1 }{ k+1 } .sin\quad \alpha \)
Hence proved.

\(\frac { sin\quad (\pi \quad cos\quad \theta ) }{ cos\quad (\pi \quad sin\quad \theta ) } =\frac { cos\quad (\pi \quad sin\quad \theta ) }{ sin\quad \left( \pi \quad sin\quad \theta \right) }\)
\(\Rightarrow cos\quad \left( \pi \quad cos\quad \theta \quad +\quad \pi \quad sin\quad \theta \right) =0\)
\(\Rightarrow cos\quad \theta \quad +\quad sin\quad \theta \quad =\quad \pm \frac { 1 }{ 2 }\)
\( \Rightarrow cos\quad \theta \quad cos\quad \frac { \pi }{ 4 } +sin\quad \theta \quad sin\quad \frac { \pi }{ 4 } =\pm \frac { 1 }{ 2\sqrt { 2 } } \)
\(\therefore \quad cos\left( \theta -\frac { \pi }{ 4 } \right) =\pm \frac { 1 }{ 2\sqrt { 2 } } \)

We have, \(cos \left( \theta +\phi \right) = m\quad cos\left( \theta -\phi \right) \)
\(\Rightarrow \frac { 1 }{ m } =\frac { cos\left( \theta -\phi \right) }{ cos\left( \theta +\phi \right) } \)
Now, \(\frac { 1-m }{ 1+m } =\frac { cos\quad \left( \theta -\phi \right) -cos\quad \left( \theta +\phi \right) }{ cos\quad \left( \theta -\phi \right) +cos\left( \theta +\phi \right) } =tan\quad \theta \quad tan\quad \phi \\ \\ \)
Ans. tan \(\theta \).

According to the given information,
we have the following

In \(\Delta \)ABC, by cosine rule,
we have
\({ AB }^{ 2 }={ AC }^{ 2 }+{ BC }^{ 2 }-2AC.BC\quad cos\frac { \pi }{ 4 } \)
\(\therefore \quad AB=\sqrt { { (250) }^{ 2 }+{ (300) }^{ 2 }-2\times 250\times 300\times \frac { 1 }{ \sqrt { 2 } } } \)
\(=\sqrt { 62500+90000-75000\sqrt { 2 } } \)
\(=\sqrt { 152500-75000\times 1.44 } \)
\(=\sqrt { 152500-108000 } =\sqrt { 44500 } =210.95m\)