Question

$\underset{\theta \to \frac{\pi }{2}}{lim}\frac{sec\theta -tan\theta }{\pi -2\theta }=$

Answer 1

We have,

${lim}_{\theta \to \frac{\pi }{2}}\frac{\sec \theta -\tan \theta }{\pi -2\theta }\frac{0}{0}form$

So,

${lim}_{\theta \to \frac{\pi }{2}}\frac{\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }}{\pi -2\theta }$

$\Rightarrow {lim}_{\theta \to \frac{\pi }{2}}\frac{1-\sin \theta }{\cos \theta (\pi -2\theta )}\frac{0}{0}form$

Applying L’ Hospital rule and we get,

$\Rightarrow {lim}_{\theta \to \frac{\pi }{2}}\frac{0-\cos \theta }{-\theta \cos \theta -\sin \theta (\pi -2\theta )}$

$\Rightarrow {lim}_{\theta \to \frac{\pi }{2}}\frac{\sin \theta }{\theta (-\sin \theta )-\cos \theta -\sin \theta (-\theta )-(\pi -2\theta )\cos \theta }$

Taking limit and we get,

$\Rightarrow \frac{\sin \frac{\pi }{2}}{\frac{\pi }{2}(-\sin \frac{\pi }{2})-\cos \frac{\pi }{2}-\sin \frac{\pi }{2}(-\frac{\pi }{2})-(\pi -2\frac{\pi }{2})\cos \frac{\pi }{2}}$

$\Rightarrow \frac{1}{-\frac{\pi }{2}-0+\frac{\pi }{2}-0}$

$\Rightarrow \frac{1}{0}=\infty$

Hence, this is the answer.