Answer on Question #62201 – Math – Trigonometry
Question
Prove that
tan ( 10 ) ⋅ tan ( 50 ) + tan ( 50 ) ⋅ tan ( 70 ) + tan ( 70 ) ⋅ tan ( 170 ) = 3. \tan(10) \cdot \tan(50) + \tan(50) \cdot \tan(70) + \tan(70) \cdot \tan(170) = 3. tan ( 10 ) ⋅ tan ( 50 ) + tan ( 50 ) ⋅ tan ( 70 ) + tan ( 70 ) ⋅ tan ( 170 ) = 3.
Solution
Let tan 10 = x \tan 10 = x tan 10 = x
tan ( 70 ) = tan ( 60 + 10 ) = tan 60 + tan 10 1 − tan 60 tan 10 = 3 + x 1 − 3 x ; \tan(70) = \tan(60 + 10) = \frac{\tan 60 + \tan 10}{1 - \tan 60 \tan 10} = \frac{\sqrt{3} + x}{1 - \sqrt{3}x}; tan ( 70 ) = tan ( 60 + 10 ) = 1 − tan 60 tan 10 tan 60 + tan 10 = 1 − 3 x 3 + x ; tan ( 50 ) = tan ( 60 − 10 ) = tan 60 − tan 10 1 + tan 60 tan 10 = 3 − x 1 + 3 x ; \tan(50) = \tan(60 - 10) = \frac{\tan 60 - \tan 10}{1 + \tan 60 \tan 10} = \frac{\sqrt{3} - x}{1 + \sqrt{3}x}; tan ( 50 ) = tan ( 60 − 10 ) = 1 + tan 60 tan 10 tan 60 − tan 10 = 1 + 3 x 3 − x ; tan ( 170 ) = tan ( 180 − 10 ) = − tan ( 10 ) = − x ; \tan(170) = \tan(180 - 10) = -\tan(10) = -x; tan ( 170 ) = tan ( 180 − 10 ) = − tan ( 10 ) = − x ; tan ( 10 ) ⋅ tan ( 50 ) + tan ( 50 ) ⋅ tan ( 70 ) + tan ( 70 ) ⋅ tan ( 170 ) = = x 3 − x 1 + 3 x + 3 − x 1 + 3 x ⋅ 3 + x 1 − 3 x − x 3 + x 1 − 3 x = = x ( 3 − x ) ( 1 − 3 x ) − ( 3 + x ) ( 1 + 3 x ) 1 − 3 x 2 + 3 − x 2 1 − 3 x 2 = = x 3 − x − 3 x + 3 x 2 − 3 − x − 3 x − 3 x 2 1 − 3 x 2 + 3 − x 2 1 − 3 x 2 = − 8 x 2 + 3 − x 2 1 − 3 x 2 = = 3 − 9 x 2 1 − 3 x 2 = 3 , when 1 − 3 x 2 ≠ 0 ⇒ x ≠ ± 1 3 . \tan(10) \cdot \tan(50) + \tan(50) \cdot \tan(70) + \tan(70) \cdot \tan(170) = \\
= x \frac{\sqrt{3} - x}{1 + \sqrt{3}x} + \frac{\sqrt{3} - x}{1 + \sqrt{3}x} \cdot \frac{\sqrt{3} + x}{1 - \sqrt{3}x} - x \frac{\sqrt{3} + x}{1 - \sqrt{3}x} = \\
= x \frac{(\sqrt{3} - x)(1 - \sqrt{3}x) - (\sqrt{3} + x)(1 + \sqrt{3}x)}{1 - 3x^2} + \frac{3 - x^2}{1 - 3x^2} = \\
= x \frac{\sqrt{3} - x - 3x + \sqrt{3}x^2 - \sqrt{3} - x - 3x - \sqrt{3}x^2}{1 - 3x^2} + \frac{3 - x^2}{1 - 3x^2} = \frac{-8x^2 + 3 - x^2}{1 - 3x^2} = \\
= \frac{3 - 9x^2}{1 - 3x^2} = 3, \text{ when } 1 - 3x^2 \neq 0 \Rightarrow x \neq \pm \frac{1}{\sqrt{3}}. tan ( 10 ) ⋅ tan ( 50 ) + tan ( 50 ) ⋅ tan ( 70 ) + tan ( 70 ) ⋅ tan ( 170 ) = = x 1 + 3 x 3 − x + 1 + 3 x 3 − x ⋅ 1 − 3 x 3 + x − x 1 − 3 x 3 + x = = x 1 − 3 x 2 ( 3 − x ) ( 1 − 3 x ) − ( 3 + x ) ( 1 + 3 x ) + 1 − 3 x 2 3 − x 2 = = x 1 − 3 x 2 3 − x − 3 x + 3 x 2 − 3 − x − 3 x − 3 x 2 + 1 − 3 x 2 3 − x 2 = 1 − 3 x 2 − 8 x 2 + 3 − x 2 = = 1 − 3 x 2 3 − 9 x 2 = 3 , when 1 − 3 x 2 = 0 ⇒ x = ± 3 1 .
Note that
tan ( 10 ) ≈ 0.6484 , tan ( 10 ∘ ) ≈ 0.1763 , 1 3 ≈ 0.5774. \tan(10) \approx 0.6484, \tan(10{}^\circ) \approx 0.1763, \frac{1}{\sqrt{3}} \approx 0.5774. tan ( 10 ) ≈ 0.6484 , tan ( 10 ∘ ) ≈ 0.1763 , 3 1 ≈ 0.5774.
Since
tan 10 ≠ ± 1 3 . \tan 10 \neq \pm \frac{1}{\sqrt{3}}. tan 10 = ± 3 1 .
Therefore, the statement is proved.
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