Question

pls solve this question i will mark as brainlist​

Answer 1

Step-by-step explanation:

LHS:

$Given:\frac{1}{secx-tanx}-\frac{1}{cosx}$

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

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

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

$=tan x$

RHS:

$Given: \frac{1}{cosx} - \frac{1}{secx + tanx}$

$= secx - [\frac{1}{secx+tanx} * \frac{secx-tanx}{secx-tanx}]$

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

$=secx -secx + tanx$

$=tanx$

∴LHS = RHS.

Hope it helps you!