Trigonometrical Equations 1 Question 25
25. Determine the smallest positive value of $x$ (in degrees) for which
$\tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan (x) \tan \left(x-50^{\circ}\right)$.
(1993, 5M)
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Answer:
Correct Answer: 25. $\frac{\sqrt{3}-1}{2}$
Solution:
- $\tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right)$
$$ \begin{aligned} & \Rightarrow \quad \frac{\tan \left(x+100^{\circ}\right)}{\tan x}=\tan \left(x+50^{\circ}\right) \tan \left(x-50^{\circ}\right) . \\ & \Rightarrow \quad \frac{\sin \left(x+100^{\circ}\right)}{\cos \left(x+100^{\circ}\right)} \cdot \frac{\cos x}{\sin x}=\frac{\sin \left(x+50^{\circ}\right) \sin \left(x-50^{\circ}\right)}{\cos \left(x+50^{\circ}\right) \cos \left(x-50^{\circ}\right)} \\ & \Rightarrow \frac{\sin \left(2 x+100^{\circ}\right)+\sin 100^{\circ}}{\sin \left(2 x+100^{\circ}\right)-\sin 100^{\circ}}=\frac{\cos 100^{\circ}-\cos 2 x}{\cos 100^{\circ}+\cos 2 x} \\ & \Rightarrow \quad\left[\sin \left(2 x+100^{\circ}\right)+\sin 100^{\circ}\right]\left[\cos 100^{\circ}+\cos 2 x\right] \\ & \quad=\left[\cos 100^{\circ}-\cos 2 x\right] \times\left[\sin \left(2 x+100^{\circ}\right)-\sin 100^{\circ}\right] \\ & \Rightarrow \quad \sin \left(2 x+100^{\circ}\right) \cdot \cos 100^{\circ}+\sin \left(2 x+100^{\circ}\right) \cdot \cos 2 x \\ & \quad+\sin 100^{\circ} \cos 100^{\circ}+\sin 100^{\circ} \cos 2 x \\ & =\cos 100^{\circ} \sin \left(2 x+100^{\circ}\right)-\cos 100^{\circ} \sin 100^{\circ} \\ & \quad-\cos 2 x \sin \left(2 x+100^{\circ}\right)+\cos 2 x \sin 100^{\circ} \end{aligned} $$
$$ \begin{array}{lr} \Rightarrow & 2 \sin \left(2 x+100^{\circ}\right) \cos 2 x+2 \sin 100^{\circ} \cos 100^{\circ}=0 \\ \Rightarrow & \sin \left(4 x+100^{\circ}\right)+\sin 100^{\circ}+\sin 200^{\circ}=0 \\ \Rightarrow & \sin \left(4 x+100^{\circ}\right)+2 \sin 150^{\circ} \cos 50^{\circ}=0 \\ \Rightarrow & \sin \left(4 x+100^{\circ}\right)+2 \cdot \frac{1}{2} \sin \left(90^{\circ}-50^{\circ}\right)=0 \\ \Rightarrow & \sin \left(4 x+100^{\circ}\right)+\sin 40^{\circ}=0 \\ \Rightarrow & \sin \left(4 x+100^{\circ}\right)=\sin \left(-40^{\circ}\right) \\ \Rightarrow & 4 x+100^{\circ}=n \pi+(-1)^{n}\left(-40^{\circ}\right) \\ \Rightarrow & 4 x=n\left(180^{\circ}\right)+(-1)^{n}\left(-40^{\circ}\right)-100^{\circ} \\ \Rightarrow & x=\frac{1}{4}\left[n\left(180^{\circ}\right)+(-1)^{n}\left(-40^{\circ}\right)-100^{\circ}\right] \end{array} $$
The smallest positive value of $x$ is obtained when $n=1$.
Therefore, $x=\frac{1}{4}\left(180^{\circ}+40^{\circ}-100^{\circ}\right)$
$$ \Rightarrow \quad x=\frac{1}{4}\left(120^{\circ}\right)=30^{\circ} $$
