Question

If $\tan \theta =1$, then find the value of $\frac{\sec \theta +\text{cosec}\theta }{\cos \theta +\sin \theta }$

Answer 1

$\tan \theta =1$

$\Rightarrow \theta =\frac{\pi }{4},\pi +\frac{\pi }{4}$

For $\theta =\frac{\pi }{4}$

$\frac{\sec \theta +\csc \theta }{\cos \theta +\sin \theta }$

$=\frac{\sec \frac{\pi }{4}+\csc \frac{\pi }{4}}{\cos \frac{\pi }{4}+\sin \frac{\pi }{4}}$

$=\frac{\sqrt{2}+\sqrt{2}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}$

$=\frac{2\sqrt{2}}{\frac{2}{\sqrt{2}}}$

$=2\sqrt{2}\times \frac{\sqrt{2}}{2}=2$

For $\theta =\pi +\frac{\pi }{4}$

$\frac{\sec \theta +\csc \theta }{\cos \theta +\sin \theta }$

$=\frac{\sec (\pi +\frac{\pi }{4})+\csc (\pi +\frac{\pi }{4})}{\cos (\pi +\frac{\pi }{4})+\sin (\pi +\frac{\pi }{4})}$

$=\frac{-\sec \frac{\pi }{4}-\csc \frac{\pi }{4}}{-\cos \frac{\pi }{4}-\sin \frac{\pi }{4}}$

$=\frac{-\sqrt{2}-\sqrt{2}}{\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}$

$=\frac{-2\sqrt{2}}{\frac{-2}{\sqrt{2}}}$

$=-2\sqrt{2}\times \frac{-\sqrt{2}}{2}=2$

$\therefore \frac{\sec \theta +\csc \theta }{\cos \theta +\sin \theta }=2$