CAIE IGCSE Additional Math 10 Trigonometry Topic Practice

$\frac{2}{\sqrt{3}}$ or $\frac{2\sqrt{3}}{3}$
B2 for $[BM=]\frac{\sqrt{3}a}{2}$ or $\sec^2 30=1+\left(\frac{1}{\sqrt{3}}\right)^2$ or $\cos^2 30=1-\left(\frac{a/2}{a}\right)^2$ oe
B1 for $[BM^2=]a^2-\frac{a^2}{4}$ soi

$\frac{2\sec x}{(\sec x-1)(\sec x+1)}$ soi
B1 for adding the two fractions.
$\frac{2\sec x}{\sec^2 x-1}$
B1 dep on previous B1 for expanding and simplifying the denominator.
$\frac{2\sec x}{\tan^2 x}$ or $\frac{2}{\tan^2 x\cos x}$
B1 dep on previous B1 for use of $\sec^2 x=1+\tan^2 x$.
$\frac{2\sec x}{\tan x}\times\cot x$ or $\frac{2\cos x}{\sin^2 x}$ oe or $\frac{2}{\tan x\sin x}$, leading to $2\operatorname{cosec}x\cot x$ AG
B1 dep on previous B1 with at least one more correct step to get to the given answer.
Alternative method:
$$\begin{aligned} \frac{1}{\frac{1}{\cos x}-1}+\frac{1}{\frac{1}{\cos x}+1}&=\frac{\cos x}{1-\cos x}+\frac{\cos x}{1+\cos x} \end{aligned}$$$=\frac{\cos x+\cos^2 x+\cos x-\cos^2 x}{1-\cos^2 x}$
(B1) dep on previous B1 for adding the two fractions.
$=\frac{2\cos x}{\sin^2 x}$
(B1) dep on previous B1 for use of $\sin^2 x=1-\cos^2 x$.
$=2\operatorname{cosec}x\times\frac{\cos x}{\sin x}=2\operatorname{cosec}x\cot x$ AG

$180^{\circ}$ or $\pi$
Allow without degrees sign

Correct curve

B1 for amplitude 3

B1 for period $180^{\circ}$

B1 for correct end points