Question

If $sec\theta =\sqrt{2}$ and $\theta$ lies in first quadrant . Find value of $\frac{1+tan\theta +cosec\theta }{1+cot\theta -cosec\theta }$

Answer 1

$\sec \theta =\sqrt{2}$

$\Rightarrow \sec \theta =\sec \frac{\pi }{4}$ $\Rightarrow \theta =\frac{\pi }{4}$

$\Rightarrow \frac{1+\tan \theta +\csc \theta }{1+\cot \theta -\csc \theta }$

$=\frac{1+\tan \frac{\pi }{4}+\csc \frac{\pi }{4}}{1+\cot \frac{\pi }{4}-\csc \frac{\pi }{4}}$

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

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

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

$=\frac{4+2+4\sqrt{2}}{4-2}$

$=\frac{6+4\sqrt{2}}{2}=3+2\sqrt{2}$