1+secA=sin²a SecA 1-cosA

{tex}\begin{aligned} & \text { To prove : } \frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sec ^{12} \mathrm{~A}}{1-\cos \mathrm{A}} \\ & \text { LHS } \frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{1+\frac{1}{\cos \mathrm{A}}}{\frac{1}{\cos \mathrm{A}}} \\ & =\frac{\frac{\cos \mathrm{A}+1}{\cos \mathrm{A}}}{\frac{1}{\cos \mathrm{A}}} \\ & =\frac{\cos \mathrm{A}+1}{\cos \mathrm{A}} \times \frac{\cos \mathrm{A}}{1} \\ & =1+\cos \mathrm{A} \\ & =1+\cos \mathrm{A} \times \frac{1-\cos \mathrm{A}}{1-\cos \mathrm{A}} \\ & =\frac{(1)^2-(\cos \mathrm{A})^2}{1-\cos \mathrm{A}} \\ & =\frac{1-\cos 2 \mathrm{~A}}{1-\cos \mathrm{A}} \end{aligned}{/tex}

{tex}\begin{aligned} & =\frac{\sin ^2 \mathrm{~A}}{1-\cos \mathrm{A}} \\ & =\mathrm{RHS} \end{aligned}{/tex}