| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard algebraic manipulation of trig fractions (finding common denominator, using cos²x + sin²x = 1) followed by a routine equation solve using the proven identity. The algebra is slightly involved but follows predictable patterns, and the equation solving is direct substitution with basic manipulation. Slightly easier than average due to the 'hence' structure guiding the solution path. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1+\sin x}{1-\sin x} - \frac{1-\sin x}{1+\sin x} \equiv \frac{(1+\sin x)^2-(1-\sin x)^2}{(1-\sin x)(1+\sin x)}\) | \*M1 | For using a common denominator of \((1-\sin x)(1+\sin x)\) and reasonable attempt at the numerator(s) |
| \(\equiv \frac{1+2\sin x+\sin^2 x-(1-2\sin x+\sin^2 x)}{(1-\sin x)(1+\sin x)}\) | DM1 | For multiplying out the numerators correctly. Condone sign errors for this mark |
| \(\equiv \frac{4\sin x}{1-\sin^2 x} \equiv \frac{4\sin x}{\cos^2 x}\) | DM1 | For simplifying denominator to \(\cos^2 x\) |
| \(\equiv \frac{4\sin x}{\cos x \cos x} \equiv \frac{4\tan x}{\cos x}\) | A1 | AG. Do not award A1 if undefined notation or missing \(x\)'s used throughout or brackets missing |
| Alternative: \(\frac{4\tan x}{\cos x} \equiv \frac{4\sin x}{\cos^2 x} \equiv \frac{4\sin x}{1-\sin^2 x}\) | \*M1 | Using \(\tan x = \frac{\sin x}{\cos x}\) and \(\cos^2 x = 1 - \sin^2 x\) |
| \(\equiv \frac{-2}{1+\sin x} + \frac{2}{1-\sin x}\) | DM1 | Separating into partial fractions |
| \(\equiv 1 + \frac{-2}{1+\sin x} + \frac{2}{1-\sin x} - 1\) | DM1 | Use of \(1\)-\(1\) or similar |
| \(\equiv -\frac{1-\sin x}{1+\sin x} + \frac{1+\sin x}{1-\sin x}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos x = \frac{1}{2}\) | \*B1 | OE. WWW |
| \(x = \frac{\pi}{3}\) | DB1 | Or AWRT 1.05 |
| \(x = 0\) from \(\tan x = 0\) or \(\sin x = 0\) | B1 | WWW. Condone extra solutions outside domain \(0\) to \(\frac{\pi}{2}\) but B0 if any inside |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1+\sin x}{1-\sin x} - \frac{1-\sin x}{1+\sin x} \equiv \frac{(1+\sin x)^2-(1-\sin x)^2}{(1-\sin x)(1+\sin x)}$ | \*M1 | For using a common denominator of $(1-\sin x)(1+\sin x)$ and reasonable attempt at the numerator(s) |
| $\equiv \frac{1+2\sin x+\sin^2 x-(1-2\sin x+\sin^2 x)}{(1-\sin x)(1+\sin x)}$ | DM1 | For multiplying out the numerators correctly. Condone sign errors for this mark |
| $\equiv \frac{4\sin x}{1-\sin^2 x} \equiv \frac{4\sin x}{\cos^2 x}$ | DM1 | For simplifying denominator to $\cos^2 x$ |
| $\equiv \frac{4\sin x}{\cos x \cos x} \equiv \frac{4\tan x}{\cos x}$ | A1 | AG. Do not award A1 if undefined notation or missing $x$'s used throughout or brackets missing |
| **Alternative:** $\frac{4\tan x}{\cos x} \equiv \frac{4\sin x}{\cos^2 x} \equiv \frac{4\sin x}{1-\sin^2 x}$ | \*M1 | Using $\tan x = \frac{\sin x}{\cos x}$ and $\cos^2 x = 1 - \sin^2 x$ |
| $\equiv \frac{-2}{1+\sin x} + \frac{2}{1-\sin x}$ | DM1 | Separating into partial fractions |
| $\equiv 1 + \frac{-2}{1+\sin x} + \frac{2}{1-\sin x} - 1$ | DM1 | Use of $1$-$1$ or similar |
| $\equiv -\frac{1-\sin x}{1+\sin x} + \frac{1+\sin x}{1-\sin x}$ | A1 | |
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos x = \frac{1}{2}$ | \*B1 | OE. WWW |
| $x = \frac{\pi}{3}$ | DB1 | Or AWRT 1.05 |
| $x = 0$ from $\tan x = 0$ or $\sin x = 0$ | B1 | WWW. Condone extra solutions outside domain $0$ to $\frac{\pi}{2}$ but B0 if any inside |
---10
\begin{enumerate}[label=(\alph*)]
\item Prove the identity $\frac { 1 + \sin x } { 1 - \sin x } - \frac { 1 - \sin x } { 1 + \sin x } \equiv \frac { 4 \tan x } { \cos x }$.
\item Hence solve the equation $\frac { 1 + \sin x } { 1 - \sin x } - \frac { 1 - \sin x } { 1 + \sin x } = 8 \tan x$ for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q10 [7]}}