Question

1-sin²x/1+cosx+1+cosx/sinx-sinx/1-cosx
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Answer 1

Answer:

$\frac{1 - \sin {}^{2} x}{1 + cosx} + \frac{1 + cosx}{sinx} - \frac{sinx}{1 - cosx} =$

$\frac{cos {}^{2}x }{1 + cosx} + \frac{0}{sinx(1 - cosx)} = \frac{cos {}^{2} x}{1 + cosx}$

$\frac{1 - {sin}^{2}x }{1 + cosx} + \frac{(1 + cosx)(1 - cosx) - sin {}^{2} x} {sinx(1 - cosx)}$

$\frac{cos {}^{2}x }{1 + cosx} + \frac{1 - cos {}^{2}x - sin {}^{2}x }{sinx(1 - cosx)} =cos^{2}x/1+cosx$

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