Trigonometry, 2ADV EQ-Bank 14 Prove that \(\sec ^2 x+\sec x\, \tan x=\dfrac{1}{1-\sin x}\). (3 marks) --- 8 WORK AREA LINES (style=lined) --- Show Answers Only \(\text{Proof (See Worked Solution)}\) Show Worked Solution \(\text {LHS }\) \(=\sec ^2 x+\sec x \, \tan x\) \(=\dfrac{1}{\cos ^2 x}+\dfrac{1}{\cos x} \cdot \dfrac{\sin x}{\cos x}\) \(=\dfrac{1+\sin x}{\cos ^2 x}\) \(=\dfrac{1+\sin x}{1-\sin ^2 x}\) \(=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}\) \(=\dfrac{1}{1-\sin x}\)