The number of elements in the set \x in [0, 180^circ] : tan(x + 100^circ) = tan(x + 50^circ) tan x tan(x - 50^circ)\$\{x \in [0, 180^{\circ}] : \tan(x + 100^{\circ}) = \tan(x + 50^{\circ}) \tan x \tan(x - 50^{\circ})\}$ is

### Related Formula textComponendo and Dividendo: fracab = fraccd implies fraca+ba-b = fracc+dc-d$$\text{Componendo and Dividendo: } \frac{a}{b} = \frac{c}{d} \implies \frac{a+b}{a-b} = \frac{c+d}{c-d}$$ sin A cos B pm cos A sin B = sin(A pm B)$$\sin A \cos B \pm \cos A \sin B = \sin(A \pm B)$$ ### Core Logic Rewrite the given equation by dividing tan x$\tan x$ to the LHS: fractan(x + 100^circ)tan x = tan(x + 50^circ) tan(x - 50^circ)$$\frac{\tan(x + 100^{\circ})}{\tan x} = \tan(x + 50^{\circ}) \tan(x - 50^{\circ})$$ Convert all terms to sines and cosines: fracsin(x + 100^circ) cos xcos(x + 100^circ) sin x = fracsin(x + 50^circ) sin(x - 50^circ)cos(x + 50^circ) cos(x - 50^circ)$$\frac{\sin(x + 100^{\circ}) \cos x}{\cos(x + 100^{\circ}) \sin x} = \frac{\sin(x + 50^{\circ}) \sin(x - 50^{\circ})}{\cos(x + 50^{\circ}) \cos(x - 50^{\circ})}$$ ### Step 1: Componendo and Dividendo Apply Componendo and Dividendo on both sides: fracsin(x + 100^circ) cos x + cos(x + 100^circ) sin xsin(x + 100^circ) cos x - cos(x + 100^circ) sin x = fracsin(x + 50^circ) sin(x - 50^circ) + cos(x + 50^circ) cos(x - 50^circ)sin(x + 50^circ) sin(x - 50^circ) - cos(x + 50^circ) cos(x - 50^circ)$$\frac{\sin(x + 100^{\circ}) \cos x + \cos(x + 100^{\circ}) \sin x}{\sin(x + 100^{\circ}) \cos x - \cos(x + 100^{\circ}) \sin x} = \frac{\sin(x + 50^{\circ}) \sin(x - 50^{\circ}) + \cos(x + 50^{\circ}) \cos(x - 50^{\circ})}{\sin(x + 50^{\circ}) \sin(x - 50^{\circ}) - \cos(x + 50^{\circ}) \cos(x - 50^{\circ})}$$ Simplify numerators and denominators using addition formulas: fracsin(2x + 100^circ)sin(100^circ) = fraccos(100^circ)-cos(2x)$$\frac{\sin(2x + 100^{\circ})}{\sin(100^{\circ})} = \frac{\cos(100^{\circ})}{-\cos(2x)}$$ Cross multiply: - sin(2x + 100^circ) cos(2x) = sin(100^circ) cos(100^circ)$$- \sin(2x + 100^{\circ}) \cos(2x) = \sin(100^{\circ}) \cos(100^{\circ})$$ ### Step 2: Trigonometric Simplification Multiply by 2: -2 sin(2x + 100^circ) cos(2x) = 2 sin(100^circ) cos(100^circ)$$-2 \sin(2x + 100^{\circ}) \cos(2x) = 2 \sin(100^{\circ}) \cos(100^{\circ})$$ Use product-to-sum formulas: - (sin(4x + 100^circ) + sin(100^circ)) = sin(200^circ)$$- (\sin(4x + 100^{\circ}) + \sin(100^{\circ})) = \sin(200^{\circ})$$ sin(4x + 100^circ) + sin(100^circ) + sin(200^circ) = 0$$\sin(4x + 100^{\circ}) + \sin(100^{\circ}) + \sin(200^{\circ}) = 0$$ Combine the constant sines: sin(100^circ) + sin(200^circ) = 2 sin(150^circ) cos(-50^circ) = 2 left(frac12right) cos(50^circ) = cos(50^circ)$$\sin(100^{\circ}) + \sin(200^{\circ}) = 2 \sin(150^{\circ}) \cos(-50^{\circ}) = 2 \left(\frac{1}{2}\right) \cos(50^{\circ}) = \cos(50^{\circ})$$ So, sin(4x + 100^circ) = -cos(50^circ)$$\sin(4x + 100^{\circ}) = -\cos(50^{\circ})$$ ### Step 3: Solving for x Convert -cos(50^circ)$-\cos(50^{\circ})$ into sine: -cos(50^circ) = -sin(40^circ) = sin(-40^circ)$$-\cos(50^{\circ}) = -\sin(40^{\circ}) = \sin(-40^{\circ})$$ Therefore: sin(4x + 100^circ) = sin(-40^circ)$$\sin(4x + 100^{\circ}) = \sin(-40^{\circ})$$ The general solution is: 4x + 100^circ = n(180^circ) + (-1)^n (-40^circ)$$4x + 100^{\circ} = n(180^{\circ}) + (-1)^n (-40^{\circ})$$ 4x = n(180^circ) + (-1)^n+1(40^circ) - 100^circ$$4x = n(180^{\circ}) + (-1)^{n+1}(40^{\circ}) - 100^{\circ}$$ x = n(45^circ) + (-1)^n+1(10^circ) - 25^circ$$x = n(45^{\circ}) + (-1)^{n+1}(10^{\circ}) - 25^{\circ}$$ ### Step 4: Finding Specific Roots Substitute integers for n$n$ to find roots in the interval x in [0, 180^circ]$x \in [0, 180^{\circ}]$: For n = 1$n = 1$: x = 45^circ + (1)(10^circ) - 25^circ = 30^circ$x = 45^{\circ} + (1)(10^{\circ}) - 25^{\circ} = 30^{\circ}$ For n = 2$n = 2$: x = 90^circ + (-1)(10^circ) - 25^circ = 55^circ$x = 90^{\circ} + (-1)(10^{\circ}) - 25^{\circ} = 55^{\circ}$ For n = 3$n = 3$: x = 135^circ + (1)(10^circ) - 25^circ = 120^circ$x = 135^{\circ} + (1)(10^{\circ}) - 25^{\circ} = 120^{\circ}$ For n = 4$n = 4$: x = 180^circ + (-1)(10^circ) - 25^circ = 145^circ$x = 180^{\circ} + (-1)(10^{\circ}) - 25^{\circ} = 145^{\circ}$ For n = 5$n = 5$: x = 225^circ + (1)(10^circ) - 25^circ = 210^circ$x = 225^{\circ} + (1)(10^{\circ}) - 25^{\circ} = 210^{\circ}$ (Out of bounds) There are exactly 4$4$ solutions in [0, 180^circ]$[0, 180^{\circ}]$. ### Pattern Recognition Fractional equality of sine and cosine products heavily points to Componendo and Dividendo. Pushing tan x$\tan x$ to the left immediately creates the standard format ready to collapse back into sin(A pm B)$\sin(A \pm B)$ and cos(A pm B)$\cos(A \pm B)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Equations