Trigonometry

Geometry Level 3

If p = tan ⁡ ( − 11 π / 6 ) , q = tan ⁡ ( 21 π / 4 ) p=\tan(-11\pi /6) ,q=\tan(21\pi /4) and r = cot ⁡ ( 283 π / 6 ) r=\cot(283\pi /6) ,then which of the following statements is/are correct?

  1. p × r = 2 p \times r = 2
  2. p,q and r are in GP.
1 only Neither 1 nor 2 Both 1 and 2 2 only

Sahil Bansal by Sahil Bansal , 21, India

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1 solution

{ p = tan ⁡ ( − 11 6 π ) = tan ⁡ ( \1 6 π ) = 1 3 q = tan ⁡ ( 21 4 π ) = tan ⁡ ( 5 1 4 π ) = tan ⁡ ( 1 4 π ) = 1 r = tan ⁡ ( 283 6 π ) = cot ⁡ ( 47 1 6 π ) = cot ⁡ ( 7 6 π ) = cot ⁡ ( 1 6 π ) = 3 \begin{cases} p = \tan{\left(-\frac{11}{6} \pi \right)} = \tan{\left(\frac{\1}{6} \pi \right)} & = \frac{1}{\sqrt{3}} \\ q = \tan{\left(\frac{21}{4} \pi \right)} = \tan{\left(5\frac{1}{4} \pi \right)} = \tan{\left(\frac{1}{4} \pi \right)} & = 1 \\ r = \tan{\left(\frac{283}{6} \pi \right)} = \cot{\left(47\frac{1}{6} \pi \right)} = \cot{\left(\frac{7}{6} \pi \right)} = \cot{\left(\frac{1}{6} \pi \right)} & = \sqrt{3} \end{cases}

{ p r = 1 3 × 3 = 1 ≠ 2 ⇒ p r = 1 = q 2 ⇒ p , q and r are in GP \begin{cases} pr = \dfrac{1}{\sqrt{3}} \times \sqrt{3} = 1 & \ne 2 \\ \Rightarrow pr = 1 = q^2 & \Rightarrow \boxed{p, q \text{ and } r \text{ are in GP}} \end{cases}

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