Question

$\frac{\mathrm{sin\theta }+\mathrm{cos\theta }}{\mathrm{sin\theta }-\mathrm{cos\theta }}+\frac{\mathrm{sin\theta }-\mathrm{cos\theta }}{\mathrm{sin\theta }+\mathrm{cos\theta }}=\frac{2}{({\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta })}=\frac{2}{(2{\sin }^2\mathrm{\theta }-1)}$

Answer 1

$\begin{array}{l}\\ \mathrm{We}\mathrm{have}\frac{\text{sin}\theta \text{+cos}\theta }{\text{sin}\theta -\text{cos}\theta }+\frac{\text{sin}\theta -\text{cos}\theta }{\text{sin}\theta \text{+cos}\theta }\\ \text{=}\frac{{\left(\text{sin}\theta \text{+cos}\theta \right)}^2+{(\text{sin}\theta -\text{cos}\theta )}^2}{(\text{sin}\theta -\text{cos}\theta )\left(\text{sin}\theta \text{+cos}\theta \right)}\\ \text{=}\frac{{\text{sin}}^2\theta +{\text{cos}}^2\theta +2\text{sin}\theta \text{cos}\theta +{\text{sin}}^2\theta {\text{+cos}}^2\theta -2\text{sin}\theta \text{cos}\theta }{{\text{sin}}^2\theta -{\text{cos}}^2\theta }\\ \text{=}\frac{1+1}{{\text{sin}}^2\theta -{\text{cos}}^2\theta }\\ \text{=}\frac{2}{{\text{sin}}^2\theta -{\text{cos}}^2\theta }\\ \text{Again,}\frac{2}{{\text{sin}}^2\theta -{\text{cos}}^2\theta }\\ \text{=}\frac{2}{{\text{sin}}^2\theta -(1-{\text{sin}}^2\theta )}\\ \text{=}\frac{2}{2{\text{sin}}^2\theta -1}\end{array}$