Tan theta + sec theta minus …
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Sia ? 6 years, 4 months ago
We have,
{tex}\mathrm { LHS } = \frac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 }{/tex}
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { ( \tan \theta + \sec \theta ) - 1 } { ( \tan \theta - \sec \theta ) + 1 }{/tex}
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { ( \sec \theta + \tan \theta ) - \left( \sec ^ { 2 } \theta - \tan ^ { 2 } \theta \right) } { \tan \theta - \sec \theta + 1 }{/tex} [{tex}\because{/tex} sec2{tex}\theta{/tex} - tan2{tex}\theta{/tex} = 1]
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { ( \sec \theta + \tan \theta ) - ( \sec \theta + \tan \theta ) ( \sec \theta - \tan \theta ) } { \tan \theta - \sec \theta + 1 }{/tex}
{tex}\Rightarrow \quad \text { LHS } = \frac { ( \sec \theta + \tan \theta ) [ 1 - ( \sec \theta - \tan \theta ) ] } { \tan \theta - \sec \theta + 1 }{/tex}
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { ( \sec \theta + \tan \theta ) ( 1 - \sec \theta + \tan \theta ) } { ( \tan \theta - \sec \theta + 1 ) }{/tex}
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { ( \sec \theta + \tan \theta ) ( \tan \theta - \sec \theta + 1 ) } { ( \tan \theta - \sec \theta + 1 ) }{/tex}
{tex}\Rightarrow \quad L H S = \sec \theta + \tan \theta = \frac { 1 } { \cos \theta } + \frac { \sin \theta } { \cos \theta } = \frac { 1 + \sin \theta } { \cos \theta }{/tex} = RHS
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