Question

Lim theta tends to π÷2 (sec theta - tan theta)

Answer 1

Using Half Angle form. for sin and cos.

Answer 2

The limit is equal to 0.

Step-by-step explanation:

Given:

$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$

Now, using trigonometric ratio,

$\sec\theta=\frac{1}{\cos\theta}\\\\\tan\theta=\frac{\sin\theta}{\cos\theta}$

$\lim_{\theta \to \frac{\pi}{2}} (\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta})\\\\\lim_{\theta \to \frac{\pi}{2}} (\frac{1-\sin\theta}{\cos\theta})$

Applying L'Hospital rule, we get:

$\lim_{\theta \to \frac{\pi}{2}} (\frac{\frac{d}{d\theta}(1-\sin\theta)}{\frac{d}{d\theta}(\cos\theta)}\\\\\lim_{\theta \to \frac{\pi}{2}}(\frac{0-\cos\theta}{-\sin\theta})\\\\\lim_{\theta \to \frac{\pi}{2}}(\frac{\cos\theta}{\sin\theta})\\\\\lim_{\theta \to \frac{\pi}{2}}(\cot\theta)..........(\because\cot\theta=\frac{\cos\theta}{\sin\theta})\\\\\cot(\frac{\pi}{2})=0$

Therefore, the value of the limit is 0.