Trigonometry (identity/ Equation)

 CIE 2020 winter Paper 22 no 11

(a) Show that $\dfrac{\sin x \tan x}{1-\cos x}=1+\sec x$.

(b) Solve the equation $5 \tan x-3 \cot x=2 \sec x$ for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.

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(a) $\begin{array}[t]{rll}\operatorname{LHS}&=\dfrac{\sin x \times \dfrac{\sin x}{\cos x}}{1-\cos x} \\&=\dfrac{1-\cos ^{2} x}{\cos x(1-\cos x)} \\&=\dfrac{(1-\cos x)(1+\cos x)}{\cos x(1-\cos x)}\\&=\dfrac{1+\cos x}{\cos x}=\sec x+1 \end{array}$

(b) $\begin{array}[t]{rl}5 \dfrac{\sin x}{\cos x}-3 \dfrac{\cos x}{\sin x}&=\dfrac{2}{\cos x} \\5 \sin ^{2} x-3\left(1-\sin ^{2} x\right)&=2 \sin x \\8 \sin ^{2} x-2 \sin x-3&=0 \\(2 \sin x+1)(4 \sin x-3)&=0 \\\sin x=-\dfrac{1}{2} \rightarrow& x=210^{\circ}, 330^{\circ} \\\sin x=\dfrac{3}{4} \rightarrow& x=48.6^{\circ}, 131.4^{\circ}\end{array} $


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