Question

Solve $\frac{d}{dx}\left(\frac{1-\cos x}{\sin x}\right)$

• A. ${\sec }^2\frac{x}{2}$
• B. $\frac{1}{2}{\sec }^2\frac{x}{2}$
• C. $2{\sec }^2\frac{x}{2}$
• D. $3{\sec }^2\frac{x}{2}$

Answer 1

Answer: (B)

$\frac{1}{2}{\sec }^2\frac{x}{2}$

We have,

$\frac{d}{dx}\left(\frac{1-\cos x}{\sin x}\right)$

$=\left(\frac{\sin x(0+\sin x)-(1-\cos x)\cos x}{{\left(\sin x\right)}^2}\right)$

$=\frac{{\sin }^{2}x-\cos x+{\cos }^{2}x}{{\left(\sin x\right)}^2}$

$=\frac{{\sin }^{2}x+{\cos }^{2}x-\cos x}{{\left(\sin x\right)}^2}$

$=\frac{1-\cos x}{{\left(\sin x\right)}^2}[\because {\sin }^{2}x+{\cos }^{2}x=1]$

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

$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

Therefore,

$=\frac{1-1+2{\sin }^2\frac{x}{2}}{{\left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)}^2}$

$=\frac{2{\sin }^2\frac{x}{2}}{4{\sin }^2\frac{x}{2}{\cos }^2\frac{x}{2}}$

$=\frac{1}{2{\cos }^2\frac{x}{2}}$

$=\frac{1}{2}{\sec }^2\frac{x}{2}$

Hence, this is the answer.