SATHEE: Trigonometrical Equations 1 Question 25

Trigonometrical Equations 1 Question 25

25. Determine the smallest positive value of $x$ (in degrees) for which

$\tan (x + 100^{\circ}) = \tan (x + 50^{\circ}) \tan (x) \tan (x - 50^{\circ})$ .

(1993, 5M)

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Answer:

Correct Answer: 25. $x = 30^{\circ}$

Solution:

$\tan (x + 100^{\circ}) = \tan (x + 50^{\circ}) \tan x \tan (x - 50^{\circ})$

$\begin{aligned} \Rightarrow \frac{\tan (x + 100^{\circ})}{\tan x} = \tan (x + 50^{\circ}) \tan (x - 50^{\circ}) . \\ \Rightarrow \frac{\sin (x + 100^{\circ})}{\cos (x + 100^{\circ})} \cdot \frac{\cos x}{\sin x} = \frac{\sin (x + 50^{\circ}) \sin (x - 50^{\circ})}{\cos (x + 50^{\circ}) \cos (x - 50^{\circ})} \\ \Rightarrow \frac{\sin (2 x + 100^{\circ}) + \sin 100^{\circ}}{\sin (2 x + 100^{\circ}) - \sin 100^{\circ}} = \frac{\cos 100^{\circ} - \cos 2 x}{\cos 100^{\circ} + \cos 2 x} \\ \Rightarrow [\sin (2 x + 100^{\circ}) + \sin 100^{\circ}] [\cos 100^{\circ} + \cos 2 x] \\ = [\cos 100^{\circ} - \cos 2 x] \times [\sin (2 x + 100^{\circ}) - \sin 100^{\circ}] \\ \Rightarrow \sin (2 x + 100^{\circ}) \cdot \cos 100^{\circ} + \sin (2 x + 100^{\circ}) \cdot \cos 2 x \\ + \sin 100^{\circ} \cos 100^{\circ} + \sin 100^{\circ} \cos 2 x \\ = \cos 100^{\circ} \sin (2 x + 100^{\circ}) - \cos 100^{\circ} \sin 100^{\circ} \\ - \cos 2 x \sin (2 x + 100^{\circ}) + \cos 2 x \sin 100^{\circ} \end{aligned}$

$\begin{array}{lr} \Rightarrow & 2 \sin (2 x + 100^{\circ}) \cos 2 x + 2 \sin 100^{\circ} \cos 100^{\circ} = 0 \\ \Rightarrow & \sin (4 x + 100^{\circ}) + \sin 100^{\circ} + \sin 200^{\circ} = 0 \\ \Rightarrow & \sin (4 x + 100^{\circ}) + 2 \sin 150^{\circ} \cos 50^{\circ} = 0 \\ \Rightarrow & \sin (4 x + 100^{\circ}) + 2 \cdot \frac{1}{2} \sin (90^{\circ} - 50^{\circ}) = 0 \\ \Rightarrow & \sin (4 x + 100^{\circ}) + \sin 40^{\circ} = 0 \\ \Rightarrow & \sin (4 x + 100^{\circ}) = \sin (- 40^{\circ}) \\ \Rightarrow & 4 x + 100^{\circ} = n π + (- 1)^{n} (- 40^{\circ}) \\ \Rightarrow & 4 x = n (180^{\circ}) + (- 1)^{n} (- 40^{\circ}) - 100^{\circ} \\ \Rightarrow & x = \frac{1}{4} [n (180^{\circ}) + (- 1)^{n} (- 40^{\circ}) - 100^{\circ}] \end{array}$

The smallest positive value of $x$ is obtained when $n = 1$ .

Therefore, $x = \frac{1}{4} (180^{\circ} + 40^{\circ} - 100^{\circ})$

$\Rightarrow x = \frac{1}{4} (120^{\circ}) = 30^{\circ}$

Trigonometrical Equations 1 Question 25

25. Determine the smallest positive value of $x$ (in degrees) for which

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Trigonometrical Equations 1 Question 25

25. Determine the smallest positive value of x (in degrees) for which

Answer:

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25. Determine the smallest positive value of $x$ (in degrees) for which