1/sec a -tan a - 1/cos …
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Sia ? 6 years, 4 months ago
We have, {tex} \frac{1}{{\sec A + \tan A}} - \frac{1}{{\cos A}} = \frac{1}{{\cos A}} - \frac{1}{{\sec A - \tan A}}{/tex}
{tex}\Rightarrow \frac{1}{{\sec A + \tan A}} + \frac{1}{{\sec A - \tan A}} = \frac{1}{{\cos A}} + \frac{1}{{\cos A}}{/tex}
LHS {tex}= \frac{1}{{\sec A + \tan A}} + \frac{1}{{\sec A - \tan A}}{/tex}
{tex}= \frac{{\sec A - \tan A + \sec A + \tan A}}{{(\sec A + \tan A)(\sec A - \tan A)}}{/tex}
{tex}= \frac{{2\sec A}}{{{{\sec }^2}A - {{\tan }^2}A}}{/tex} {tex} \left[ {\because (a + b)(a - b) = ({a^2} - {b^2}} \right]{/tex}
{tex}= \frac{{2\sec A}}{1}{/tex} {tex} \left[ {\because {{\sec }^2}A - {{\tan }^2}A = 1} \right]{/tex}
= 2secA
RHS {tex}= \frac{1}{{\cos A}} + \frac{1}{{\cos A}}{/tex}
{tex}= \frac{{1 + 1}}{{\cos A}}{/tex}
{tex}= \frac{2}{{\cos A}}{/tex}
= 2 secA
LHS = RHS
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