Question

$\frac{{\cos }^3\mathrm{\theta }+{\sin }^3\mathrm{\theta }}{\mathrm{cos\theta }+\mathrm{sin\theta }}+\frac{{\cos }^3\mathrm{\theta }-{\sin }^3\mathrm{\theta }}{\mathrm{cos\theta }-\mathrm{sin\theta }}=2$

Answer 1

$\begin{array}{l}\\ \text{LHS=}\frac{{\text{cos}}^3\theta {\text{+sin}}^3\theta }{\text{cos}\theta \text{+sin}\theta }+\frac{{\text{cos}}^3\theta -{\text{sin}}^3\theta }{\text{cos}\theta -\text{sin}\theta }\\ \text{=}\frac{\left(\text{cos}\theta \text{+sin}\theta \right)({\text{cos}}^2\theta -\text{cos}\theta \text{sin}\theta {\text{+sin}}^2\theta )}{\left(\text{cos}\theta \text{+sin}\theta \right)}+\frac{(\text{cos}\theta -\text{sin}\theta )({\text{cos}}^2\theta +\text{cos}\theta \text{sin}\theta {\text{+sin}}^2\theta )}{(\text{cos}\theta -\text{sin}\theta )}\\ \\ \text{=}({\text{cos}}^2\theta {\text{+sin}}^2\theta -\text{cos}\theta \text{sin}\theta )+({\text{cos}}^2\theta {\text{+sin}}^2\theta +\text{cos}\theta \text{sin}\theta )\\ \text{=}(1-\text{cos}\theta \text{sin}\theta )+(1+\text{cos}\theta \text{sin}\theta )\\ \text{=2}\\ \text{=RHS}\end{array}$
Hence, LHS= RHS