Question

Prove the following.

(ii) $\sqrt{\frac{\cos ecx-1}{\cos ecx+1}}=\frac{1}{secx+\tan x}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sqrt{\frac{\cos ecx-1}{\cos ecx+1}} \\ =\sqrt{\frac{{\displaystyle \frac{1}{\sin x}}-1}{{\displaystyle \frac{1}{\sin x}}+1}} \\ =\sqrt{\frac{1-\sin x}{1+\sin x}} \\ =\sqrt{\frac{\left(1-\sin x\right)\left(1+\sin x\right)}{\left(1+\sin x\right)\left(1+\sin x\right)}} \\ =\sqrt{\frac{1-{\sin }^{2}x}{{\left(1+\sin x\right)}^2}} \\ =\sqrt{\frac{\cos {}^{2}x}{{\left(1+\sin x\right)}^2}} \end{gathered}$$

$$\begin{gathered} =\frac{\cos X}{1+\sin X} \\ =\frac{{\displaystyle \frac{\cos X}{\cos X}}}{{\displaystyle \frac{1+\sin X}{\cos X}}} \\ =\frac{1}{{\displaystyle \frac{1}{\cos X}}+{\displaystyle \frac{\sin X}{\cos X}}} \\ =\frac{1}{secX+\tan X} \\ =\mathrm{RHS} \end{gathered}$$

Hence, proved.