| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Substitution t equals tan |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring the standard t-substitution formulas (sec θ = (1+t²)/(1-t²), tan θ = 2t/(1-t²)) and algebraic manipulation. Part (a) is methodical substitution and simplification; part (b) cleverly uses the result from (a) with a reciprocal relationship. While requiring more steps than typical A-level questions and being from Further Maths, it follows a well-established technique without requiring novel insight, placing it moderately above average difficulty. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sight of \(\sec\theta = \frac{1+t^2}{1-t^2}\) and \(\tan\theta = \frac{2t}{1-t^2}\) at least once each | B1 | Uses correct identities at least once each; may be seen in (a) or (b) |
| \(\frac{29 - 21\sec\theta}{20 - 21\tan\theta} = \frac{29 - 21\left(\frac{1+t^2}{1-t^2}\right)}{20 - 21\left(\frac{2t}{1-t^2}\right)}\) | M1 | Substitutes their identities into the equation |
| \(= \frac{29(1-t^2)-21(1+t^2)}{20(1-t^2)-21(2t)} = \frac{8-50t^2}{-20t^2-42t+20}\) | M1 | Multiplies numerator and denominator by \(1-t^2\) and simplifies to a quadratic with all terms collected |
| \(= \frac{-2(5t-2)(5t+2)}{-2(5t-2)(2t+5)} = \frac{5t+2}{2t+5}\) * cso | A1\* | Cancels factors to achieve given expression, with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{20+21\tan\theta}{29+21\sec\theta} = \frac{20+21\left(\frac{2t}{1-t^2}\right)}{29+21\left(\frac{1+t^2}{1-t^2}\right)} = \frac{20(1-t^2)+21(2t)}{29(1-t^2)+21(1+t^2)}\) | M1 | Substitutes into LHS and multiplies numerator and denominator by \(1-t^2\) |
| \(\frac{20+42t-20t^2}{50-8t^2} = \frac{2(5-2t)(5t+2)}{2(5-2t)(5+2t)} = \ldots\) | M1 | Simplifies to quadratic, collects terms, factorises and cancels in LHS to match expressions |
| Achieves \(= \frac{5t+2}{2t+5}\) then concludes LHS = RHS (or equivalent conclusion) | A1\* | Achieves correct expressions for both sides and gives conclusion that result is true |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sight of $\sec\theta = \frac{1+t^2}{1-t^2}$ and $\tan\theta = \frac{2t}{1-t^2}$ at least once each | **B1** | Uses correct identities at least once each; may be seen in (a) or (b) |
| $\frac{29 - 21\sec\theta}{20 - 21\tan\theta} = \frac{29 - 21\left(\frac{1+t^2}{1-t^2}\right)}{20 - 21\left(\frac{2t}{1-t^2}\right)}$ | **M1** | Substitutes their identities into the equation |
| $= \frac{29(1-t^2)-21(1+t^2)}{20(1-t^2)-21(2t)} = \frac{8-50t^2}{-20t^2-42t+20}$ | **M1** | Multiplies numerator and denominator by $1-t^2$ and simplifies to a quadratic with all terms collected |
| $= \frac{-2(5t-2)(5t+2)}{-2(5t-2)(2t+5)} = \frac{5t+2}{2t+5}$ * cso | **A1\*** | Cancels factors to achieve given expression, with no errors seen |
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## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{20+21\tan\theta}{29+21\sec\theta} = \frac{20+21\left(\frac{2t}{1-t^2}\right)}{29+21\left(\frac{1+t^2}{1-t^2}\right)} = \frac{20(1-t^2)+21(2t)}{29(1-t^2)+21(1+t^2)}$ | **M1** | Substitutes into LHS and multiplies numerator and denominator by $1-t^2$ |
| $\frac{20+42t-20t^2}{50-8t^2} = \frac{2(5-2t)(5t+2)}{2(5-2t)(5+2t)} = \ldots$ | **M1** | Simplifies to quadratic, collects terms, factorises and cancels in LHS to match expressions |
| Achieves $= \frac{5t+2}{2t+5}$ then concludes LHS = RHS (or equivalent conclusion) | **A1\*** | Achieves correct expressions for both sides and gives conclusion that result is true |\begin{enumerate}
\item (a) Use $t = \tan \frac { \theta } { 2 }$ to show that, where both sides are defined
\end{enumerate}
$$\frac { 29 - 21 \sec \theta } { 20 - 21 \tan \theta } \equiv \frac { 5 t + 2 } { 2 t + 5 }$$
(b) Hence, again using $t = \tan \frac { \theta } { 2 }$, prove that, where both sides are defined
$$\frac { 20 + 21 \tan \theta } { 29 + 21 \sec \theta } \equiv \frac { 29 - 21 \sec \theta } { 20 - 21 \tan \theta }$$
\hfill \mbox{\textit{Edexcel FP1 AS 2022 Q3 [7]}}