Sec theta minus tan theta upon …

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Sia ? 6 years, 4 months ago
We have,
{tex} \mathrm { LHS } = \frac { \sec \theta - \tan \theta } { \sec \theta + \tan \theta }{/tex}
{tex} \Rightarrow \quad \mathrm { LHS } = \frac { \sec \theta - \tan \theta } { \sec \theta + \tan \theta } \times \frac { \sec \theta - \tan \theta } { \sec \theta - \tan \theta }{/tex}
{tex} \Rightarrow \quad \mathrm { LHS } = \frac { ( \sec \theta - \tan \theta ) ^ { 2 } } { \sec ^ { 2 } \theta - \tan ^ { 2 } \theta } = \frac { ( \sec \theta - \tan \theta ) ^ { 2 } } { 1 }{/tex} [{tex} \because{/tex} sec2{tex} \theta{/tex} - tan2{tex} \theta{/tex} = 1]
{tex} \Rightarrow \quad \mathrm { LHS } = \sec ^ { 2 } \theta - 2 \sec \theta \tan \theta + \tan ^ { 2 } \theta{/tex}
{tex} \Rightarrow \quad \mathrm { LHS } = \left( 1 + \tan ^ { 2 } \theta \right) - 2 \sec \theta \tan \theta + \tan ^ { 2 } \theta{/tex} [{tex} \because{/tex} sec2{tex} \theta{/tex} = 1 + tan2{tex} \theta{/tex}]
{tex} \Rightarrow{/tex} LHS = 1 - 2sec{tex} \theta{/tex} tan{tex} \theta{/tex} + 2tan2{tex} \theta{/tex} = RHS
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