Question

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

Answer 1

1/(secx-tanx) - 1/cosx
= secx + tanx -secx
= secx - (secx-tanx)
= 1/cosx - (sec²x-tan²x)/(secx+tanx)
=1/cosx - 1/(secx+tanx)

hence proved

Answer 2

Answer:

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

Step-by-step explanation:

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

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

= $\frac{secx+tanx}{sec^{2}x-tan^{2}x}-secx$

=$secx+tanx-secx$

/* Since , sec²x - tan²x = 1 */

=$secx -(secx-tanx)$

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

=$secx - \frac{sec^{2}x-tan^{2}x}{(secx-tanx}$

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

= $RHS$

Therefore,

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

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