Given, m sin\(\theta \)  = nsin (\(\theta \) + 2\(\alpha \))
 \(\Rightarrow \) \(\frac { m }{ n } = \frac { sin\left( \theta +2\alpha \right) }{ sin\theta } \)
Applying componedo and dividendo rule, we get
\(\frac { m+n }{ m-n } =\quad \frac { sin\left( \theta +2\alpha \right) +sin\theta }{ sin\left( \theta +2\alpha \right) -sin\theta } \)
\(=\frac { 2sin\left( \frac { \theta +2\alpha +\theta }{ 2 } \right) cos\left( \frac { \theta +2\alpha +\theta }{ 2 } \right) }{ 2cos\left( \frac { \theta +2\alpha +\theta }{ 2 } \right) sin\left( \frac { \theta +2\alpha +\theta }{ 2 } \right) } \)
\(\left[ sinA+sinB=2sin\left( \frac { A+B }{ 2 } \right) cos\left( \frac { A-B }{ 2 } \right) and\quad sinA-sinB=2cos\left( \frac { A+B }{ 2 } \right) sin\left( \frac { A-B }{ 2 } \right) \right]\)
\(=\frac { 2sin(\theta +\alpha )\quad cos\alpha }{ 2cos(\theta +\alpha )\quad cos\alpha } =\quad tan(\theta +\alpha )cot\alpha\)
Hence proved.

Given, a cos A=b cos B
\(\Rightarrow \) k sin A cos A=k sin B cos B
\(\Rightarrow \) sin 2A=sin 2B
\(\Rightarrow \)  sin 2A - sin 2B=0
\(\Rightarrow \)  2cos(A+B)sin(A-B)=0
\(\Rightarrow \)  cos(A+B)=0 or sin(A-B)=0
\(\Rightarrow \)  A+B=\(\frac { \pi }{ 2 } \) or A-B=0
Ans. \(\angle \)C=\(\frac { \pi }{ 2 } \) or A=B

\(Let\frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } } =K\)
\(\Rightarrow a=K\sin { A,b=K\sin { B\quad and } } c=K\sin { C } \)
\(Now,\frac { \sin { \left( A-B \right) } }{ \sin { \left( A+B \right) } } =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } =\frac { { K }^{ 2 }\sin { ^{ 2 }A-{ K }^{ 2 }\sin { ^{ 2 }B } } }{ { K }^{ 2 }\sin { ^{ 2 }A+{ K }^{ 2 }\sin { ^{ 2 }B } } } \)
\( \Rightarrow \quad \quad \frac { \sin { \left( A-B \right) } }{ \sin { \left( A+B \right) } } =\frac { \sin { ^{ 2 }A-\sin { ^{ 2 }B } } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } \)
\(\Rightarrow \frac { \sin { \left( A-B \right) } }{ \sin { \left( \pi -C \right) } } =\frac { \sin { \left( A+B \right) } \sin { \left( A-B \right) } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } \quad \left[ \therefore A+B=\pi -C \right] \)
\(\Rightarrow \frac { \sin { \left( A-B \right) } }{ \sin { C } } =\frac { \sin { C\sin { \left( A-B \right) } } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } \)
\(\Rightarrow \frac { \sin { \left( A-B \right) } }{ \sin { C } } =\frac { \sin { C\sin { \left( A-B \right) } } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } =0\)
\(\Rightarrow \sin { A-B\left[ \frac { 1 }{ \sin { C } } -\frac { \sin { C } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } \right] } =0\)
\(\Rightarrow \sin { \left( A-B \right) =0 } \quad or\quad \frac { 1 }{ \sin { C } } -\frac { \sin { C } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } =0\)
\(\Rightarrow A=B\quad or\quad \sin { ^{ 2 }A+\sin { ^{ 2 }B } -\sin { ^{ 2 }C } } =0\)
\( \Rightarrow A=B\quad or\quad \frac { { a }^{ 2 } }{ { k }^{ 2 } } +\frac { { b }^{ 2 } }{ { k }^{ 2 } } -\frac { { c }^{ 2 } }{ { k }^{ 2 } } =0\)
\(\Rightarrow A=B\quad or\quad { a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\)
\(\therefore \) Either the triangle is isosceles or right angled.
Hence proved.

LHS = \(tan\quad \theta \quad tan\quad ({ 60 }^{ o }-\theta )tan\quad ({ 60 }^{ o }+\theta )\)
\(=\frac { sin\quad \theta }{ cos\quad \theta } .\frac { sin\quad \left( { 60 }^{ o }-\theta \right) }{ cos\quad \left( { 60 }^{ o }-\theta \right) } .\frac { sin\quad \left( { 60 }^{ o }+\theta \right) }{ cos\quad \left( { 60 }^{ o }+\theta \right) } \)
\(=\frac { sin\quad \theta [2sin\quad \left( { 60 }^{ o }-\theta \right) sin\quad \left( { 60 }^{ o }+\theta \right) ] }{ cos\quad \theta [2cos\quad \left( { 60 }^{ o }-\theta \right) cos\quad \left( { 60 }^{ o }+\theta \right) ] } \)
\(=\frac { sin\quad \theta (cos\quad 2\theta \quad -\quad cos\quad { 120 }^{ o }) }{ cos\quad \theta (cos\quad { 120 }^{ o }+cos\quad 2\theta ) } \)
\(\because\)  2 sinA sinB = cos(A−B)−cos(A+B) and 2 cosA cosB = cos(A+B)+cos(A−B)]
\(=\frac { sin\quad \theta \left( cos\quad 2\theta +\frac { 1 }{ 2 } \right) }{ cos\quad \theta \quad \left( cos\quad 2\theta -\frac { 1 }{ 2 } \right) } \left[ \because cos { 120 }^{ o }= cos \left( { 180 }^{ o }- { 60 }^{ o } \right) = -cos\quad { 60 }^{ o }=-\frac { 1 }{ 2 } \right] \)
\(=\frac { sin\quad \theta cos\quad 2\theta +\frac { 1 }{ 2 } sin\theta }{ cos\quad \theta \quad cos\quad 2\theta -\frac { 1 }{ 2 } cos\theta } =\frac { 2sin\quad \theta cos\quad 2\theta +sin\theta }{ 2cos\quad \theta \quad cos\quad 2\theta -cos\quad \theta }\)
\(=\frac { sin\quad \left( \theta +2\theta \right) +sin\quad (-\theta )+sin\theta }{ cos\quad (\theta +2\theta )+cos\quad (-\theta )-cos\quad \theta } \)
\(\left[ \therefore \quad 2sin\quad A\quad cos\quad B\quad =\quad sin\quad (A+B)+sin\quad (A-B)\quad and\quad 2cos\quad A\quad cos\quad B\quad =\quad cos\quad (A+B)+cos\quad (A-B) \right] \)
\(=\frac { sin\quad 3\theta -sin\quad \theta +sin\quad \theta }{ cos\quad 3\theta +cos\quad \theta -cos\quad \theta } \quad \quad \left[ \because sin\quad (-\theta )=-sin\theta \quad and\quad cos\quad (-\theta )=cos\quad \theta \right] \)
\(=\frac { sin\quad 3\theta }{ cos\quad 3\theta } =tan\quad 3\theta =RHS\)
Hence proved.

Given, co s\(\alpha \)+cos\(\beta\) = 0 and sin \(\alpha \)+sin\(\beta\) = 0
on squaring both equations, we get
(cos\(\alpha \)+cos\(\beta\) )² =0 ...(i)
and  ( sin \(\alpha \)+sin\(\beta\))² = 0 ....(ii)
On subtracting Eq.(ii) from Eq.(i), we get
(cos\(\alpha \)+cos\(\beta\) )² +( sin \(\alpha \)+sin\(\beta\))² =0
\(\Rightarrow \)(cos² \(\alpha \)+cos² \(\beta\)+2cos\(\alpha \) cos \(\beta\) - (sin²\(\alpha \)+ sin² \(\beta\)+2sin\(\alpha \) sin\(\beta\))=0  [(a+b)²= a²+b²+2ab]
\(\Rightarrow \)cos² \(\alpha \)+cos² \(\beta\)+2cos\(\alpha \) cos \(\beta\)-sin²\(\alpha \) - sin²\(\beta\)-2sin \(\alpha \)+sin\(\beta\)= 0 
\(\Rightarrow \)(cos² \(\alpha \) - sin² \(\alpha \)) + (cos² \(\beta\) - sin² \(\beta\)) + 2[cos² \(\alpha \)+cos² \(\beta\) - 2sin \(\alpha \) sin\(\beta\)]= 0   
\(\left[\begin{array}{l} \because \cos 2 x=\cos ^{2} x-\sin ^{2} x \text { and } \\ \cos A \cos B-\sin A \sin B=\cos (A+B) \end{array}\right]\)
\(\therefore \quad \cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta)\)