Question

If $\sec \theta =\sqrt{2}$ and $\frac{3\pi }{2}<\theta <2\pi$, if value of $\frac{1+\tan \theta +\csc \theta }{1+\cot \theta -\csc \theta }=m$ then find $-m$.

Answer 1

We have: $sec\theta =\sqrt{2}\Rightarrow cos\theta =\frac{1}{\sqrt{2}}$

$\therefore sin\theta =\pm \sqrt{1-co{s}^2\theta }=\pm \sqrt{1-\frac{1}{2}}=\pm \frac{1}{\sqrt{2}}$

But $\theta$ lies in the third quadrant in which $sin\theta$ is negative.

$\therefore sin\theta =\frac{-1}{\sqrt{2}}\iff cosec\theta =-\sqrt{2}$and $tan\theta =\frac{sin\theta }{cos\theta }\Rightarrow tan\theta =-\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{1}=-1$

$\Rightarrow cot\theta =-1$

$\therefore \frac{1+tan\theta +cosec\theta }{1+tan\theta -cosec\theta }=\frac{1-1-\sqrt{2}}{1-1+\sqrt{2}}=-1$