Question

$\frac{\mathrm{sin\theta }}{(\mathrm{sec\theta }+\mathrm{tan\theta }-1)}+\frac{\mathrm{cos\theta }}{(\mathrm{cosec\theta }+\mathrm{cot\theta }-1)}=1$

Answer 1

$\begin{array}{l}\\ \text{LHS=}\frac{\text{sin}\theta }{(\text{sec}\theta \text{+tan}\theta -1)}+\frac{\text{cos}\theta }{(\text{cosec}\theta \text{+cot}\theta -1)}\\ \text{=}\frac{\text{sin}\theta \text{cos}\theta }{1+\text{sin}\theta -\text{cos}\theta }+\frac{\text{cos}\theta \text{sin}\theta }{1+\text{cos}\theta -\text{sin}\theta }\\ \text{=sin}\theta \text{cos}\theta [\frac{1}{1+(\text{sin}\theta -\text{cos}\theta )}+\frac{1}{1-(\text{sin}\theta -\text{cos}\theta )}]\\ \text{=sin}\theta \text{cos}\theta \left[\frac{1-(\text{sin}\theta -\text{cos}\theta )+1+(\text{sin}\theta -\text{cos}\theta )}{\{1+(\text{sin}\theta -\text{cos}\theta )\}\{1-(\text{sin}\theta -\text{cos}\theta )\}}\right]\\ \text{=sin}\theta \text{cos}\theta \left[\frac{1-\text{sin}\theta \text{+cos}\theta \text{+1+sin}\theta -\text{cos}\theta }{1-{(\text{sin}\theta -\text{cos}\theta )}^2}\right]\\ \text{=}\frac{2\text{sin}\theta \text{cos}\theta }{1-({\text{sin}}^2\theta +{\text{cos}}^2\theta -2\text{sin}\theta \text{cos}\theta )}\\ \text{=}\frac{2\text{sin}\theta \text{cos}\theta }{2\text{sin}\theta \text{cos}\theta }\\ \text{=1}\\ \text{=RHS}\end{array}$
Hence, LHS = RHS