June 2022 Paper 1 Q21
21 Show that \(\qquad \dfrac{4\cos^2 x + 3\sin^2 x - 4}{\cos^2 x} \equiv -\tan^2 x\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| LHS Use of: \(\quad \cos^2 x \equiv 1 - \sin^2 x\) or \(\quad \sin^2 x \equiv 1 - \cos^2 x\) or \(\quad 3\sin^2 x + 3\cos^2 x \equiv 3\) in numerator to get: \(4(1 - \sin^2 x) + 3\sin^2 x - 4\) or \(4\cos^2 x + 3(1 - \cos^2 x) - 4\) or \(3 + \cos^2 x - 4\) | M1 | oe must be used as part of a solution (nothing for just stating it) |
| LHS \(\dfrac{4 - 4\sin^2 x + 3\sin^2 x - 4}{\cos^2 x}\) or simplification of the other forms leading to \(\dfrac{\cos^2 x - 1}{\cos^2 x}\) | M1dep | one step away from the A mark this could imply the first M1 provided they have stated the identity used from the list in the first M mark |
| \(-\dfrac{\sin^2 x}{\cos^2 x} \equiv -\tan^2 x\) | A1 | oe |
| Alternative method 2 | ||
| LHS \(\dfrac{\left[4\cos^2 x + 3\sin^2 x - 4\left(\cos^2 x + \sin^2 x\right)\right]}{\cos^2 x}\) | M1 | |
| \(\dfrac{\left[4\cos^2 x + 3\sin^2 x - 4\cos^2 x - 4\sin^2 x\right]}{\cos^2 x}\) | M1 | |
| \(-\dfrac{\sin^2 x}{\cos^2 x} \equiv -\tan^2 x\) | A1 | oe |
| Alternative method 3 | ||
| RHS \(-\tan^2 x \equiv -\dfrac{\sin^2 x}{\cos^2 x}\) | M1 | |
| \(\dfrac{\left[4\left(\sin^2 x + \cos^2 x\right) - 4 - \sin^2 x\right]}{\cos^2 x}\) | M1 | |
| \(\dfrac{\left[4\cos^2 x + 3\sin^2 x - 4\right]}{\cos^2 x}\) | A1 | |
Additional guidance
| Either starts with the left and finishes with the right or vice versa. Max M2 for any working that meets in the middle by trying to solve an equation Only mark using one of the alts – once the candidate starts to treat the solution as an equation by moving terms around from one side of the \(\equiv\) to the other then stop awarding marks The exception to this would be if a candidate uses identities to manipulate the LHS to an expression correctly and also then manipulates the RHS correctly to the same expression. They would then need to state that these two manipulations show the LHS \(\equiv\) RHS |