Question

(tanx)/(secx-1 )-(sinx)/(1+cosx)

Answer 1

Hi friend
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Your answer
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$\frac{ \tan(x) }{( \sec(x) - 1 )} - \frac{ \sin(x) }{(1 + \cos(x))} \\ \\ = > \frac{ \tan(x) (1 + cosx) - sinx(secx - 1)}{(secx - 1)(cosx + 1)} \\ \\ = > \frac{ \frac{sinx}{cosx} \times 1 + \frac{sinx}{cosx} \times cosx - sinx \times \frac{1}{cosx} + sinx}{( \frac{1}{cosx} - 1)(cosx + 1)} \\ \\ = > \frac{sinx + sinx + \frac{sinx}{cosx} - \frac{sinx}{cosx}}{ \frac{(1 - cosx)(1 + cosx)}{cosx} } \\ \\ = > \frac{2sinx}{ \frac{1 - cos {}^{2}x }{cosx} } \\ \\ = > 2sinx \times \frac{cosx}{sin {}^{2}x } \\ \\ = > 2 \times \frac{cosx}{sinx} \\ \\ = > 2cotx$


HOPE IT HELPS

Answer 2

Heya mate
here is your answer

I hope this will help you...✌️✌️
thank u:-)