| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Prove identity then solve |
| Difficulty | Moderate -0.3 Part (a) is a routine trigonometric identity proof requiring standard manipulation of sec, tan, sin, cos with algebraic simplification. Part (b) uses the proven identity to reduce to a quadratic in tan x, then solve within a restricted domain. This is a standard two-part question with straightforward techniques and no novel insight required, making it slightly easier than average. |
| 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 |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)\left(\frac{1}{\sin x} + 1\right)\) | B1 | Uses "\(\tan x = \sin x \div \cos x\)" throughout |
| \(\left(\frac{1-\sin x}{\cos x}\right)\left(\frac{1+\sin x}{\sin x}\right)\) or \(\left(\frac{1 - \sin^2 x}{\cos x \sin x}\right)\) | M1 | Correct algebra leading to two or four terms |
| \(\left(\frac{\cos^2 x}{\cos x \sin x}\right)\) | A1 | OE. A correct expression which can be cancelled directly to \(\frac{\cos x}{\sin x}\), e.g. \(\frac{\cos x(1-\sin^2 x)}{\sin x(1-\sin^2 x)}\) |
| \(\left(\frac{\cos^2 x}{\cos x \sin x}\right) = \left(\frac{\cos x}{\sin x}\right) = \frac{1}{\tan x}\) | A1 | AG. Must show cancelling. If \(x\) is missing throughout their working withhold this mark. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Uses (a) \(\rightarrow \frac{1}{\tan x} = 2\tan^2 x\), \(\tan^3 x = \frac{1}{2}\) | M1 | Reducing to \(\tan^3 x = k\). |
| \((x =)\ 38.4°\) | A1 | AWRT. Ignore extra answers outside the range \(0\) to \(180°\) but A0 if within. |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)\left(\frac{1}{\sin x} + 1\right)$ | B1 | Uses "$\tan x = \sin x \div \cos x$" throughout |
| $\left(\frac{1-\sin x}{\cos x}\right)\left(\frac{1+\sin x}{\sin x}\right)$ or $\left(\frac{1 - \sin^2 x}{\cos x \sin x}\right)$ | M1 | Correct algebra leading to two or four terms |
| $\left(\frac{\cos^2 x}{\cos x \sin x}\right)$ | A1 | OE. A correct expression which can be cancelled directly to $\frac{\cos x}{\sin x}$, e.g. $\frac{\cos x(1-\sin^2 x)}{\sin x(1-\sin^2 x)}$ |
| $\left(\frac{\cos^2 x}{\cos x \sin x}\right) = \left(\frac{\cos x}{\sin x}\right) = \frac{1}{\tan x}$ | A1 | AG. Must show cancelling. If $x$ is missing throughout their working withhold this mark. |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Uses **(a)** $\rightarrow \frac{1}{\tan x} = 2\tan^2 x$, $\tan^3 x = \frac{1}{2}$ | M1 | Reducing to $\tan^3 x = k$. |
| $(x =)\ 38.4°$ | A1 | AWRT. Ignore extra answers outside the range $0$ to $180°$ but A0 if within. |6
\begin{enumerate}[label=(\alph*)]
\item Prove the identity $\left( \frac { 1 } { \cos x } - \tan x \right) \left( \frac { 1 } { \sin x } + 1 \right) \equiv \frac { 1 } { \tan x }$.
\item Hence solve the equation $\left( \frac { 1 } { \cos x } - \tan x \right) \left( \frac { 1 } { \sin x } + 1 \right) = 2 \tan ^ { 2 } x$ for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q6 [6]}}