@avn bha , My answer is quite long. So please try to find any shorter way (if any) and do tell me...
Dividing the given eqn. by \(cos{ \theta }\). You get the equation in terms of \(\tan { \theta } \quad and\quad \sec { \theta } \). Squaring both sides and converting \(\sec ^{ 2 }{ \theta } =1+\tan ^{ 2 }{ \theta } \) and simplifying, u get an equation, quadratic in \(\tan { \theta } \) which looks like

\(4{ a }^{ 2 }{ b }^{ 2 }\tan ^{ 2 }{ \theta } -4ab({ a }^{ 2 }-{ b }^{ 2 })\tan { \theta } +{ ({ a }^{ 2 }-{ b }^{ 2 }) }^{ 2 }=0\\ Using\quad quadratic\quad formula\quad you\quad get\quad \\ \tan { \theta } =\frac { 4ab({ a }^{ 2 }-{ b }^{ 2 }) }{ 8{ a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ 2ab } \)

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TopNewesthow did u get it?

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@avn bha , My answer is quite long. So please try to find any shorter way (if any) and do tell me... Dividing the given eqn. by \(cos{ \theta }\). You get the equation in terms of \(\tan { \theta } \quad and\quad \sec { \theta } \). Squaring both sides and converting \(\sec ^{ 2 }{ \theta } =1+\tan ^{ 2 }{ \theta } \) and simplifying, u get an equation, quadratic in \(\tan { \theta } \) which looks like

\(4{ a }^{ 2 }{ b }^{ 2 }\tan ^{ 2 }{ \theta } -4ab({ a }^{ 2 }-{ b }^{ 2 })\tan { \theta } +{ ({ a }^{ 2 }-{ b }^{ 2 }) }^{ 2 }=0\\ Using\quad quadratic\quad formula\quad you\quad get\quad \\ \tan { \theta } =\frac { 4ab({ a }^{ 2 }-{ b }^{ 2 }) }{ 8{ a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ 2ab } \)

CHEERS!!!:)

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\(\tan \theta = \frac{a^{2}-b^{2}}{2ab}\)

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