Question

show that 1/secx - tanx - 1/cosx = 1/cosx - 1/secx + tanx

Answer 1

Answer:

$\frac{1}{secx-tanx}-\frac{1}{cosx}=\frac{1}{cosx}-\frac{1}{secx+tanx}$

Step-by-step explanation:

WKT

$sec^2x-tan^2x=1\\\\(secx-tanx)(secx+tanx)=1$

Now

$\frac{1}{secx-tanx}-\frac{1}{cosx}\\\\secx=\frac{1}{cosx}\\\\\frac{secx+tanx}{(secx-tanx)(secx+tanx)}-\frac{1}{cosx}\\\\secx+tanx-secx\\\\secx-(secx-tanx)\\\\\frac{1}{cosx}-\frac{(secx-tanx)(secx+tanx)}{secx+tanx}\\\\\frac{1}{cosx}-\frac{1}{secx+tanx}$

Hope It Helps

Answer 2

$\frac{1}{secx - tanx} - \frac{1}{cosx} \\ \\ = secx + tanx - secx \\ \\ = secx - (secx - tanx) \\ \\ = \frac{1}{cosx} - \frac{1}{secx + tanx}$