Question

10 x by sec x minus 1 minus sin x by 1 + cos x ​

Answer 1

Answer:

$\frac{tanx}{secx - 1} - \frac{sinx}{1 + cosx} \\ take \: the \: lcm \: in \: denomintor \\ \frac{tanx(1 + cosx) - sinx(secx - 1)}{(secx - 1)(1 + cosx)}$

$\frac{tanx + sinx - tanx + sinx}{secx + 1 - 1 - cosx} \\ \frac{2sinx}{ \frac{1}{cosx} - cosx } \: \\ \frac{2sinx}{ \frac{1 - {cos}^{2}x }{cosx} } = \frac{2sinx.cosx}{ {sin}^{2}x } = \: 2cotx$

thus 2 cotx is answer..

hope this will help you..