Question

(i) $\frac{1+\mathrm{cos\theta }+\mathrm{sin\theta }}{1+\mathrm{cos\theta }-\mathrm{sin\theta }}=\frac{1+\mathrm{sin\theta }}{\mathrm{cos\theta }}$
(ii) $\frac{\mathrm{sin\theta }+1-\mathrm{cos\theta }}{\mathrm{cos\theta }-1+\mathrm{sin\theta }}=\frac{1+\mathrm{sin\theta }}{\mathrm{cos\theta }}$

Answer 1

$\begin{array}{l}\\ \left(\text{i}\right)\text{LHS=}\frac{1+\text{cos}\theta \text{+sin}\theta }{1+\text{cos}\theta -\text{sin}\theta }\\ \text{=}\frac{\{(1+\text{cos}\theta )+\text{sin}\theta \}\{(1+\text{cos}\theta )+\text{sin}\theta \}}{\{(1+\text{cos}\theta )-\text{sin}\theta \}\{(1+\text{cos}\theta )+\text{sin}\theta \}}\text{{Multiplying the numerator and denominator by (}1+\text{cos}\theta +\text{sinθ)}}\\ \text{=}\frac{{\{(1+\text{cos}\theta )+\text{sin}\theta \}}^2}{\{{(1+\text{cos}\theta )}^2-{\text{sin}}^2\theta \}}\\ \text{=}\frac{1+{\text{cos}}^2\theta +2\text{cos}\theta {\text{+sin}}^2\theta +2\text{sin}\theta (1+\text{cos}\theta )}{1+{\text{cos}}^2\theta +2\text{cos}\theta -{\text{sin}}^2\theta }\\ \text{=}\frac{2+2\text{cos}\theta \text{+}2\text{sin}\theta (1+\text{cos}\theta )}{1+{\text{cos}}^2\theta +2\text{cos}\theta -(1-{\text{cos}}^2\theta )}\\ \text{=}\frac{2(1+\text{cos}\theta )+2\text{sin}\theta (1+\text{cos}\theta )}{2{\text{cos}}^2\theta +2\text{cos}\theta }\\ \text{=}\frac{2(1+\text{cos}\theta )(1+\text{sin}\theta )}{2\text{cos}\theta (1+\text{cos}\theta )}\\ \text{=}\frac{1+\text{sin}\theta }{\text{cos}\theta }\\ \text{=RHS}\\ \left(\text{ii}\right)\text{LHS=}\frac{\text{sin}\theta \text{+1}-\text{cos}\theta }{\text{cos}\theta -1+\text{sin}\theta }\\ \text{=}\frac{(\text{sin}\theta \text{+1}-\text{cos}\theta )\left(\text{sin}\theta \text{+cos}\theta \text{+1}\right)}{(\text{cos}\theta -1+\text{sin}\theta )\left(\text{sin}\theta \text{+cos}\theta \text{+1}\right)}\{\mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{by}1+\text{cos}\theta +\text{sinθ }}\\ \text{=}\frac{{(\text{sin}\theta +1)}^2-{\text{cos}}^2\theta }{{\left(\text{sin}\theta \text{+cos}\theta \right)}^2-1^2}\\ \text{=}\frac{{\text{sin}}^2\theta +1+2\text{sin}\theta -{\text{cos}}^2\theta }{{\text{sin}}^2\theta +{\text{cos}}^2\theta +2\text{sin}\theta \text{cos}\theta -1}\\ \text{=}\frac{{\text{sin}}^2\theta +{\text{sin}}^2\theta +{\text{cos}}^2\theta +2\text{sin}\theta -{\text{cos}}^2\theta }{2\text{sin}\theta \text{cos}\theta }\\ \text{=}\frac{2{\text{sin}}^2\theta +2\text{sin}\theta }{2\text{sin}\theta \text{cos}\theta }\\ \text{=}\frac{2\text{sin}\theta (1+\text{sin}\theta )}{\text{2sin}\theta \text{cos}\theta }\\ \text{=}\frac{\text{1+sin}\theta }{\text{cos}\theta }\\ \text{=RHS}\end{array}$