| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (i) requires recognizing that 1 - sin²x = cos²x and simplifying to tan²x - 1, which is straightforward algebraic manipulation with standard identities. Part (ii) uses this result to form a quadratic in tan x, then solving tan x = 2 or -3 in the given range. This is a standard C2 identity proof followed by routine equation solving, slightly easier than average due to the clear structure and hint provided. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\sin^2 x - \cos^2 x}{1 - \sin^2 x} = \frac{\sin^2 x - \cos^2 x}{\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} = \tan^2 x - 1\) | M1 | Use either \(\sin^2 x + \cos^2 x = 1\), or \(\tan x = \frac{\sin x}{\cos x}\) |
| A1 | Use other identity to obtain given answer convincingly |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan^2 x - 1 = 5 - \tan x\) | B1 | State correct equation |
| \(\tan^2 x + \tan x - 6 = 0\) | ||
| \((\tan x - 2)(\tan x + 3) = 0\) | M1 | Attempt to solve three term quadratic in \(\tan x\) |
| \(\tan x = 2, \tan x = -3\) | A1 | Obtain 2 and \(-3\) as roots of their quadratic |
| \(x = 63.4°, 243°\) \(\quad\) \(x = 108°, 288°\) | M1 | Attempt to solve \(\tan x = k\) (at least one root) |
| A1 ft | Obtain at least 2 correct roots | |
| A1 | Obtain all 4 correct roots |
## Question 7:
### Part (i)
| $\frac{\sin^2 x - \cos^2 x}{1 - \sin^2 x} = \frac{\sin^2 x - \cos^2 x}{\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} = \tan^2 x - 1$ | M1 | Use either $\sin^2 x + \cos^2 x = 1$, or $\tan x = \frac{\sin x}{\cos x}$ |
| | A1 | Use other identity to obtain given answer convincingly |
### Part (ii)
| $\tan^2 x - 1 = 5 - \tan x$ | B1 | State correct equation |
| $\tan^2 x + \tan x - 6 = 0$ | | |
| $(\tan x - 2)(\tan x + 3) = 0$ | M1 | Attempt to solve three term quadratic in $\tan x$ |
| $\tan x = 2, \tan x = -3$ | A1 | Obtain 2 and $-3$ as roots of their quadratic |
| $x = 63.4°, 243°$ $\quad$ $x = 108°, 288°$ | M1 | Attempt to solve $\tan x = k$ (at least one root) |
| | A1 ft | Obtain at least 2 correct roots |
| | A1 | Obtain all 4 correct roots |
---7 (i) Show that $\frac { \sin ^ { 2 } x - \cos ^ { 2 } x } { 1 - \sin ^ { 2 } x } \equiv \tan ^ { 2 } x - 1$.\\
(ii) Hence solve the equation
$$\frac { \sin ^ { 2 } x - \cos ^ { 2 } x } { 1 - \sin ^ { 2 } x } = 5 - \tan x$$
for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{OCR C2 2010 Q7 [8]}}