June 2017 Paper 2 Q23
23 In this question, \(\tan x \neq 0\) and \(\sin x \neq 0\)
Show that \(\quad \dfrac{1}{\tan^2 x} - \dfrac{1}{\sin^2 x} \quad\) is a constant. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Replaces \(\tan x\) with \(\dfrac{\sin x}{\cos x}\) at least once in given expression | M1 | eg \(\dfrac{1}{\frac{\sin^2 x}{\cos^2 x}} - \dfrac{1}{\sin^2 x}\) |
| Correct steps leading to the single fraction \(\dfrac{\cos^2 x - 1}{\sin^2 x}\) or \(\dfrac{\cos^2 x - 1}{1 - \cos^2 x}\) or \(\dfrac{1 - \sin^2 x - 1}{\sin^2 x}\) or \(\dfrac{\cos^2 x - \cos^2 x - \sin^2 x}{\sin^2 x}\) or \(\dfrac{-\sin^2 x}{\sin^2 x}\) | M1dep | |
| \(\dfrac{\cos^2 x - 1}{\sin^2 x} = \dfrac{-\sin^2 x}{\sin^2 x} = -1\) or \(\dfrac{\cos^2 x - 1}{1 - \cos^2 x} = -1\) or \(\dfrac{1 - \sin^2 x - 1}{\sin^2 x} = -1\) or \(\dfrac{-\sin^2 x}{\sin^2 x} = -1\) | A1 | Must see all steps leading to \(-1\) |
| Alternative method 2 | ||
| Replaces \(\tan x\) with \(\dfrac{\sin x}{\cos x}\) at least once in given expression | M1 | eg \(\dfrac{\sin^2 x - \frac{\sin^2 x}{\cos^2 x}}{\sin^2 x \, \frac{\sin^2 x}{\cos^2 x}}\) |
| Correct steps leading to the single fraction \(\dfrac{\sin^2 x(\cos^2 x - 1)}{\sin^4 x}\) or \(\dfrac{-\sin^4 x}{\sin^4 x}\) | M1dep | |
| \(\dfrac{\sin^2 x(\cos^2 x - 1)}{\sin^4 x} = \dfrac{-\sin^4 x}{\sin^4 x} = -1\) or \(\dfrac{-\sin^4 x}{\sin^4 x} = -1\) | A1 | Must see all steps leading to \(-1\) |
Additional guidance
| Allow \(\cos^2\theta\) etc or \(\cos^2\) or c\(^2\) or \((\cos x)^2\) for \(\cos^2 x\) etc | |
| Condone \(\cos x^2\) for \(\cos^2 x\) etc | |
| Only substituting values for \(x\) | Zero |
| \(\dfrac{\cos^2 x - 1}{\sin^2 x}\) etc with no working | Zero |
| Alt 2 \(\quad \dfrac{\sin^2 x\cos^2 x - \sin^2 x}{\sin^4 x}\) with no further working | M1M0A0 |
| Any fully correct response that shows how the given expression is equal to \(-1\) is awarded 3 marks eg \(\dfrac{1}{\frac{\sin^2 x}{\cos^2 x}} - \dfrac{1}{\sin^2 x} = \dfrac{\cos^2 x}{\sin^2 x} - \dfrac{1}{\sin^2 x} = \dfrac{1 - \sin^2 x}{\sin^2 x} - \dfrac{1}{\sin^2 x}\) \(= \dfrac{1}{\sin^2 x} - \dfrac{\sin^2 x}{\sin^2 x} - \dfrac{1}{\sin^2 x} = -1\) | 3 marks |
| \(\cot^2 x - \operatorname{cosec}^2 x = -1\) | 3 marks |