Question

Evaluate: $1-\frac{{\sin }^{2}x}{1+\cos x}+\frac{1+\cos x}{\sin x}-\frac{\sin x}{1-\cos x}=$

• A. $\sin x$
• B. $\cos x$
• C. $cosecx$
• D. $\sec x$

Answer 1

Answer: (B)

$\cos x$

$1-\frac{{\sin }^{2}x}{1+\cos x}+\frac{1+\cos x}{\sin x}-\frac{\sin x}{1-\cos x}$

$\Rightarrow \frac{(1+\cos x)-(1-{\cos }^{2}x)}{(1+\cos x)}+\frac{1-{\cos }^{2}x-{\sin }^{2}x}{\sin x(1-\cos x)}$

$\Rightarrow \frac{(1+\cos x)(1-1+\cos x)}{1+\cos x}+\frac{1-1}{\sin x(1-\cos x)}$

$\Rightarrow \cos x+0$

$\Rightarrow \cos x$.