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EEJ MAIN Mathematics QUESTION #7433
Question 1
The expression \(\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A}\) can be written as:
  • sinA cosA + 1
  • secA cosecA + 1✔️
  • tanA + cotA
  • secA + cosecA
Correct Answer Explanation
Let's simplify: \(\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A} = \frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}} = \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}} = \frac{\sin^2 A}{\cos A(\sin A - \cos A)} + \frac{\cos^2 A}{\sin A(\cos A - \sin A)} = \frac{\sin^2 A}{\cos A(\sin A - \cos A)} - \frac{\cos^2 A}{\sin A(\sin A - \cos A)} = \frac{\sin^3 A - \cos^3 A}{\sin A \cos A (\sin A - \cos A)} = \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A \cos A (\sin A - \cos A)} = \frac{1 + \sin A \cos A}{\sin A \cos A} = \sec A \csc A + 1\).