Question

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

• A. $2tanx$
• B. $2sinx$
• C. $6cosx$
• D. $2cotx$

Answer 1

The correct option is D $2cotx$

$\frac{tanx}{secx-1}-\frac{sinx}{1+cosx}$
$=\frac{\frac{sinx}{cosx}}{\frac{1-cosx}{cosx}}-\frac{sinx}{1+cosx}$
$=sin(\frac{1}{1-cosx}-\frac{1}{1+cosx})$
$=sinx\left(\frac{2.cosx}{1-co{s}^{2}x}\right)$
$=\frac{2.sinx.cosx}{si{n}^{2}x}=2cotx$