Question

Prove:

$\frac{{\text{cos}}^2\theta }{(1-\text{tan}\theta )}+\frac{{\text{sin}}^3\theta }{(\text{sin}\theta -\text{cos}\theta )}=(1+\text{sin}\theta \text{cos}\theta )$

Answer 1

$LHS=\frac{{\text{cos}}^2\theta }{(1-\text{tan}\theta )}+\frac{{\text{sin}}^3\theta }{(\text{sin}\theta -\text{cos}\theta )}$

$=\frac{{\text{cos}}^2\theta }{(1-\frac{\text{sin}\theta }{\text{cos}\theta })}+\frac{{\text{sin}}^3\theta }{(\text{sin}\theta -\text{cos}\theta )}$ [$\because \text{tan}\theta =\frac{\text{sin}\theta }{\text{cos}\theta }$]

$=\frac{{\text{cos}}^3\theta }{(\text{cos}\theta -\text{sin}\theta )}+\frac{{\text{sin}}^3\theta }{(\text{sin}\theta -\text{cos}\theta )}$

$=\frac{1}{(\text{cos}\theta -\text{sin}\theta )}[{\text{cos}}^3\theta -{\text{sin}}^3\theta ]$

$=\frac{1}{(\text{cos}\theta -\text{sin}\theta )}\left[\right({\text{cos}}^2\theta +\text{cos}\theta \text{sin}\theta +{\text{sin}}^2\theta \left)\right(\text{cos}\theta -\text{sin}\theta \left)\right]$ [$\because (a^3-b^3)=(a-b)(a^2+ab+b^2)$]

$=[{\text{sin}}^2\theta +{\text{cos}}^2\theta +\text{cos}\theta \text{sin}\theta ]$

$=(1+\text{sin}\theta \text{cos}\theta )$ [$\because {\text{sin}}^2\theta +{\text{cos}}^2\theta =1$ ]

$=RHS$

Hence, $\frac{{\text{cos}}^2\theta }{(1-\tan \theta )}+\frac{{\text{sin}}^3\theta }{(\text{sin}\theta -\text{cos}\theta )}=(1+\text{sin}\theta \text{cos}\theta )$