Question

Prove that 1/(secA-TanA) - 1/CosA = 1/CosA - 1/(SecA+TanA). ​

Answer 1

$\huge\color{red}{\underline{\underline {Question}}}$

Prove that :-

$\frac{1}{secA - tanA} - \frac{1}{cosA} = \frac{1}{cosA} - \frac{1}{secA + tanA}$

$\huge\color{red}{\underline{\underline {Answer}}}$

$Given, \\ \\ \frac{1}{secA - tanA} - \frac{1}{cosA} \: = \frac{1}{cosA} - \frac{1}{secA + tanA} \\ \\ L.H.S= \frac{1}{secA - tanA} - \frac{1}{cosA} \\ \\ It \: can \: also \: be \: written \: as \\ \\ \frac{1}{secA - tanA} + \frac{1}{secA + tanA} \\ \\ Similarly , \\ \\ R. H. S = \frac{1}{cosA} - \frac{1}{secA + tanA} \\ \\ It \: can \: also \: be \: written \: as \\ \\ \frac{1}{cosA} + \frac{1}{cosA} = \frac{2}{cosA} \\ \\ Now \: taking \: L.H.S= \: \frac{1}{secA - tanA} + \frac{1}{secA + tanA} \\ \\ = > \frac{secA + tanA + secA - tanA}{(secA - tanA)(secA + tanA)} \\ \\ = > \frac{2}{cosA} = R. H. S$

$\mathtt{\bf{\large{\underline {\red{\: \: \:Hence,\: proved.}}}}}$

Answer 2

$\huge\color{red}{\underline{\underline {Question}}}$

Prove that :-

$\frac{1}{secA - tanA} - \frac{1}{cosA} = \frac{1}{cosA} - \frac{1}{secA + tanA}$

$\huge\color{red}{\underline{\underline {Answer}}}$

$Given, \\ \\ \frac{1}{secA - tanA} - \frac{1}{cosA} \: = \frac{1}{cosA} - \frac{1}{secA + tanA} \\ \\ L.H.S= \frac{1}{secA - tanA} - \frac{1}{cosA} \\ \\ It \: can \: also \: be \: written \: as \\ \\ \frac{1}{secA - tanA} + \frac{1}{secA + tanA} \\ \\ Similarly , \\ \\ R. H. S = \frac{1}{cosA} - \frac{1}{secA + tanA} \\ \\ It \: can \: also \: be \: written \: as \\ \\ \frac{1}{cosA} + \frac{1}{cosA} = \frac{2}{cosA} \\ \\ Now \: taking \: L.H.S= \: \frac{1}{secA - tanA} + \frac{1}{secA + tanA} \\ \\ = > \frac{secA + tanA + secA - tanA}{(secA - tanA)(secA + tanA)} \\ \\ = > \frac{2}{cosA} = R. H. S$

$\mathtt{\bf{\large{\underline {\red{\: \: \:Hence,\: proved.}}}}}$