Prove that tan A + sec …
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Posted by Adarsh Roy 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
{tex}\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = \frac{{\tan A + \sec A - ({{\sec }^2}A - {{\tan }^2}A)}}{{\tan A - \sec A + 1}}{/tex} [{tex}\because{/tex} 1 + tan2A = sec2A {tex}\Rightarrow{/tex} sec2A - tan2A = 1]
{tex} = \frac{{\tan A + \sec A - \{ (\sec A - \tan A)(\sec A + \tan A)\} }}{{\tan A - \sec A + 1}}{/tex}
Now, take (tanA + secA) as a common term, we get
{tex} = \frac{{(\tan A + \sec A)(1 - \sec A + \tan A)}}{{\tan A - \sec A + 1}}{/tex}
= tanA + secA
{tex} = \frac{{\sin A}}{{\cos A}} + \frac{1}{{\cos A}}{/tex}
{tex} = \frac{{1 + \sin A}}{{\cos A}}{/tex}
{tex}\therefore{/tex} Hence proved
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