Standard +0.8 Part (a) is a routine identity proof requiring basic manipulation of reciprocal trig functions. Part (b) is more challenging: it requires recognizing that the identity from (a) implies tan(x) = tan(3x - π/9), then solving a compound angle equation with multiple solutions in the given domain, requiring careful consideration of the tangent's periodicity.
12. a. Show that
$$\sec \theta - \cos \theta = \sin \theta \tan \theta \quad \theta \neq ( \pi n ) ^ { 0 } \quad n \in Z$$
b. Hence, or otherwise, solve for \(0 < x \leq \pi\)
$$\sec x - \cos x = \sin x \tan \left( 3 x - \frac { \pi } { 9 } \right)$$
States or uses \(\sec \theta = \frac{1}{\cos\theta}\)
M1
Attempts to obtain a single fraction
A1
Shows all the necessary steps leading to given answer
Part b: Hence, or otherwise, solve for \(0 < x \le \pi\): \(\sec x - \cos x = \sin x \tan(3x - \frac{\pi}{9})\)
Answer
Marks
Guidance
\(\sin x \tan x = \sin x \tan\left(3x - \frac{\pi}{9}\right)\)
\(\tan x = \tan\left(3x - \frac{\pi}{9}\right)\)
\(x = 3x - \frac{\pi}{9} \Rightarrow x = \frac{\pi}{18}\)
M1
Uses part (a), cancels or factorises out the \(\sin x\) term, to establish that one solution is found when \(x = 3x - \frac{\pi}{9}\)
Second solution can be found from \(x + \pi = 3x - \frac{\pi}{9} \Rightarrow x = \frac{5\pi}{9}\)
A1
\(x = \frac{\pi}{18}\)
Third solution can be found from \(\sin x = 0 \Rightarrow x = \pi\)
M1
Second solution can be found by solving \(x + \pi = 3x - \frac{\pi}{9}\)
A1
\(x = \frac{5\pi}{9}\)
B1
Deduces that a solution can be found from \(\sin x = 0 \Rightarrow x = \pi\)
(3) marks + (5) marks = (8) marks total for Question 12
**Part a:** Show that $\sec \theta - \cos \theta = \sin \theta \tan \theta$
| $\frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2\theta}{\cos \theta} = \frac{\sin^2\theta}{\cos\theta} = \sin\theta\tan\theta$ | B1 | States or uses $\sec \theta = \frac{1}{\cos\theta}$ |
| --- | --- | --- |
| | M1 | Attempts to obtain a single fraction |
| | A1 | Shows all the necessary steps leading to given answer |
**Part b:** Hence, or otherwise, solve for $0 < x \le \pi$: $\sec x - \cos x = \sin x \tan(3x - \frac{\pi}{9})$
| $\sin x \tan x = \sin x \tan\left(3x - \frac{\pi}{9}\right)$ | | |
| --- | --- | --- |
| $\tan x = \tan\left(3x - \frac{\pi}{9}\right)$ | | |
| $x = 3x - \frac{\pi}{9} \Rightarrow x = \frac{\pi}{18}$ | M1 | Uses part (a), cancels or factorises out the $\sin x$ term, to establish that one solution is found when $x = 3x - \frac{\pi}{9}$ |
| Second solution can be found from $x + \pi = 3x - \frac{\pi}{9} \Rightarrow x = \frac{5\pi}{9}$ | A1 | $x = \frac{\pi}{18}$ |
| Third solution can be found from $\sin x = 0 \Rightarrow x = \pi$ | M1 | Second solution can be found by solving $x + \pi = 3x - \frac{\pi}{9}$ |
| | A1 | $x = \frac{5\pi}{9}$ |
| | B1 | Deduces that a solution can be found from $\sin x = 0 \Rightarrow x = \pi$ |
**(3) marks + (5) marks = (8) marks total for Question 12**
---