Question

Prove that $\frac{\text{cos}x}{(1-sinx)}=tan(\frac{\pi }{4}+\frac{x}{2})$

Answer 1

We have LHS = $\frac{\text{cos}x}{(1-sinx)}$

=$\frac{({\text{cos}}^2\frac{x}{2}-{\text{sin}}^2\frac{x}{2})}{({\text{cos}}^2\frac{x}{2}+{\text{sin}}^2\frac{x}{2}-2\text{sin}\frac{x}{2}\text{cos}\frac{x}{2})}$

$\left[\begin{array}{cc}\because \text{cos}x=({\text{cos}}^2\frac{x}{2}-\text{sin}\frac{x}{2},){\text{cos}}^2\frac{x}{2}+{\text{sin}}^2\frac{x}{2}=1& \\ \text{and sin}x=2\text{sin}\frac{x}{2}\text{cos}\frac{x}{2}& \end{array}\right]$

=$\frac{(\text{cos}\frac{x}{2}-\text{sin}\frac{x}{2})(\text{cos}\frac{x}{2}+\text{sin}\frac{x}{2})}{{(\text{cos}\frac{x}{2}-\text{sin}\frac{x}{2})}^2}$

$=\frac{(\text{cos}\frac{x}{2}+\text{sin}\frac{x}{2})}{(\text{cos}\frac{x}{2}-\text{sin}\frac{x}{2})}=\frac{(1+\text{tan}\frac{x}{2})}{(1-\text{tan}\frac{x}{2})}=\frac{(\text{tan}\frac{\pi }{4}+\text{tan}\frac{x}{2})}{(1-\text{tan}\frac{\pi }{4}.\text{tan}\frac{x}{2})}$

$\left[\text{dividing num. and denom. by cos}\frac{x}{2}\right]$

=$\text{tan}(\frac{x}{4}+\frac{x}{2})$ = RHS