The smallest positive value of x (in degrees) for which tanx+100∘=tanx+50∘tanx⋅tanx−50∘ is

⇒ tan⁡x+100∘tan⁡x+50∘=tan⁡x+50∘tan⁡x∘⇒ sin⁡x+100∘cos⁡x−50∘cos⁡x+100∘sin⁡x−50∘=sin⁡x+50∘sin⁡xcos⁡x+50∘cos⁡x⇒sin⁡x+100∘cos⁡x−50∘+cos⁡x+100∘sin⁡x−50∘sin⁡x+100∘cos⁡x−50∘−cos⁡x+100∘sin⁡x−50∘=sin⁡x+50∘sin⁡x+cos⁡x+50∘cos⁡xsin⁡x+50∘sin⁡x−cos⁡x+50∘cos⁡x⇒ sin⁡x+100∘+x−50∘sin⁡x+100∘−x+50∘=cos⁡x+50∘−x−cos⁡x+50∘+x⇒ sin⁡2x+50∘sin⁡150∘=cos⁡50∘−cos⁡2x+50∘⇒sin⁡2x+50∘cos⁡2x+50∘=−sin⁡150∘cos⁡50∘⇒2sin⁡2x+50∘cos⁡2x+50∘=−2cos⁡60∘cos⁡50∘ ∵sin⁡150∘=cos⁡60∘⇒sin⁡4x+100∘=sin⁡270−50∘⇒sin⁡4x+100∘=sin⁡220∘⇒4x+100∘=220∘⇒x=30∘

The smallest positive value of x (in degrees) for which tanx+100∘=tanx+50∘tanx⋅tanx−50∘ is

⇒ tan⁡x+100∘tan⁡x+50∘=tan⁡x+50∘tan⁡x∘⇒ sin⁡x+100∘cos⁡x−50∘cos⁡x+100∘sin⁡x−50∘=sin⁡x+50∘sin⁡xcos⁡x+50∘cos⁡x⇒sin⁡x+100∘cos⁡x−50∘+cos⁡x+100∘sin⁡x−50∘sin⁡x+100∘cos⁡x−50∘−cos⁡x+100∘sin⁡x−50∘=sin⁡x+50∘sin⁡x+cos⁡x+50∘cos⁡xsin⁡x+50∘sin⁡x−cos⁡x+50∘cos⁡x⇒ sin⁡x+100∘+x−50∘sin⁡x+100∘−x+50∘=cos⁡x+50∘−x−cos⁡x+50∘+x⇒ sin⁡2x+50∘sin⁡150∘=cos⁡50∘−cos⁡2x+50∘⇒sin⁡2x+50∘cos⁡2x+50∘=−sin⁡150∘cos⁡50∘⇒2sin⁡2x+50∘cos⁡2x+50∘=−2cos⁡60∘cos⁡50∘ ∵sin⁡150∘=cos⁡60∘⇒sin⁡4x+100∘=sin⁡270−50∘⇒sin⁡4x+100∘=sin⁡220∘⇒4x+100∘=220∘⇒x=30∘

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$The smallest positive value of x (in degrees) for which \tan (x + 100^{\circ}) = \tan (x + 50^{\circ}) \tan x \cdot \tan (x - 50^{\circ}) is$

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Detailed Solution

$\begin{array}{l} \Rightarrow \frac{\tan (x + 100^{\circ})}{\tan (x + 50^{\circ})} = \tan (x + 50^{\circ}) \tan x^{\circ} \\ \Rightarrow \frac{\sin (x + 100^{\circ}) \cos (x - 50^{\circ})}{\cos (x + 100^{\circ}) \sin (x - 50^{\circ})} = \frac{\sin (x + 50^{\circ}) \sin x}{\cos (x + 50^{\circ}) \cos x} \\ \Rightarrow \frac{\sin (x + 100^{\circ}) \cos (x - 50^{\circ}) + \cos (x + 100^{\circ}) \sin (x - 50^{\circ})}{\sin (x + 100^{\circ}) \cos (x - 50^{\circ}) - \cos (x + 100^{\circ}) \sin (x - 50^{\circ})} \\ = \frac{\sin (x + 50^{\circ}) \sin x + \cos (x + 50^{\circ}) \cos x}{\sin (x + 50^{\circ}) \sin x - \cos (x + 50^{\circ}) \cos x} \\ \Rightarrow \frac{\sin (x + 100^{\circ} + x - 50^{\circ})}{\sin (x + 100^{\circ} - x + 50^{\circ})} = \frac{\cos (x + 50^{\circ} - x)}{- \cos (x + 50^{\circ} + x)} \\ \Rightarrow \frac{\sin (2 x + 50^{\circ})}{\sin 150^{\circ}} = \frac{\cos 50^{\circ}}{- \cos (2 x + 50^{\circ})} \\ \begin{array}{cc} \Rightarrow & \sin (2 x + 50^{\circ}) \cos (2 x + 50^{\circ}) = - \sin 150^{\circ} \cos 50^{\circ} \\ \Rightarrow & 2 \sin (2 x + 50^{\circ}) \cos (2 x + 50^{\circ}) = - 2 \cos 60^{\circ} \cos 50^{\circ} \\ [∵ \sin 150^{\circ} = \cos 60^{\circ}] \\ \Rightarrow & \sin (4 x + 100^{\circ}) = \sin (270 - 50^{\circ}) \\ \Rightarrow & \sin (4 x + 100^{\circ}) = \sin 220^{\circ} \\ \Rightarrow & 4 x + 100^{\circ} = 220^{\circ} \Rightarrow x = 30^{\circ} \end{array} \end{array}$