# Trigonometry (Identity/ Equation)

(CIE 0606/2021 March/12/Q8)
(a) (i) Show that $\sin x\tan x+\cos x=\sec x$. [3]

(ii) Hence solve the equation $\sin\dfrac{\theta}{2}\tan\dfrac{\theta}{2}+\cos\dfrac{\theta}{2}=4,$ for $0\le \theta\le4\pi$,  where $\theta$ is in radians. [4]

(b) Solve the equation $\cot(y+38^{\circ})=\sqrt{3}$ for $0\le y\le 360^{\circ}.$ [3]

(a) (i) $\begin{array}[t]{rll}\sin x \tan x+\cos x &=\sin x \cdot \dfrac{\sin x}{\cos x}+\cos x \\&=\dfrac{\sin ^{2} x+\cos ^{2} x}{\cos x} \\&=\dfrac{1}{\cos x}=\sec x . \end{array}$

(a)(ii) $4=\sin\left(\dfrac{\theta}{2}\right) \tan\left(\dfrac{\theta}{2}\right)+\cos\left(\dfrac{\theta}{2}\right)=\sec\left(\dfrac{\theta}{2}\right)$

$\begin{array}[t]{rll}\therefore \cos \dfrac{\theta}{2}&=\dfrac 14\\\dfrac{\theta}{2}&=1.3181 \text { or } \dfrac{\theta}{2}=2 \pi-1.3181=4.9651 \\\theta&= 2.64 \text { or } \theta=9.93\end{array}$

(b)  $\begin{array}[t]{rll} \cot \left(y+38^{\circ}\right)&=\sqrt{3}\\ \tan\left(y+38^{\circ}\right)&=\dfrac{1}{\sqrt 3}\\ \left(y+38^{\circ}\right)&=30^{\circ} \text{ (reiect) or} 172^{\circ} \text{ or }352^{\circ}\\ \therefore y&=172^{\circ} \text{ or } 352^{\circ}\end{array}$