Prove that: 1- = ( 4 + 2 )

Question

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

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Answer

$\text{LHS}=\frac{\cos\text{x}}{1-\sin\text{x}}$ $\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{\sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}-2\sin\frac{\text{x}}{2},\frac{\cos\text{x}}{2}}$ $[\because\cos2\text{x}=\cos^2\text{x}-\sin2\text{x}\ \&\sin^2\text{x}+\cos^2\text{x}]=1$ $\frac{(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2})(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2})}{(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2})}$ $=\frac{\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}}$ Dividing numeretor and denominator by $\cos\frac {\text{x}}{2}$ $=\frac{1+\tan\frac{\text{x}}{2}}{1=\tan\frac{\text{x}}{2}}$ $\tan(\frac{\pi}{4}+\frac{\text{x}}{2})=\text{RHS}$

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