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Value of \(\cot\left(\dfrac{\pi}{20}\right)\cot\left(\dfrac{3\pi}{20}\right)\cot\left(\dfrac{5\pi}{20}\right)\cot\left(\dfrac{7\pi}{20}\right)\cot\left(\dfrac{9\pi}{20}\right)\)

This question was previously asked in
HTET 2025 Level 1 PRT Question Paper (5-Jul-2026)
The correct answer is

1

Convert each angle to degrees: \(\dfrac{\pi}{20}=9^\circ\), \(\dfrac{3\pi}{20}=27^\circ\), \(\dfrac{5\pi}{20}=45^\circ\), \(\dfrac{7\pi}{20}=63^\circ\), \(\dfrac{9\pi}{20}=81^\circ\).

So the product is \(\cot 9^\circ\cot 27^\circ\cot 45^\circ\cot 63^\circ\cot 81^\circ\).

Note \(\cot 45^\circ=1\). Also, using \(\cot(90^\circ-\theta)=\tan\theta\): \(\cot 81^\circ=\cot(90^\circ-9^\circ)=\tan 9^\circ\) and \(\cot 63^\circ=\cot(90^\circ-27^\circ)=\tan 27^\circ\).

So the product becomes \((\cot 9^\circ\tan 9^\circ)\times(\cot 27^\circ\tan 27^\circ)\times 1=1\times1\times1=1\).

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Important Questions from Trigonometric Identities

  1. What is \(\rm \frac{1+tan^2\theta}{1+cot^2\theta}-\left(\frac{1-tan\theta}{1-cot\theta}\right)^2\) equal to?

  2. If 3sin θ + 5cos θ = 5, then the value of 5sin θ - 3cos θ is equal to: 

  3. If angle C of a triangle ABC is a right angle where a, b and c are the sides opposite to the angles A, B and C respectively then what is tan A + tan B equal to?

  4. If \(\sin \left( {A - B} \right) = \frac{1}{2}\)  and  \(\cos \left( {A + B} \right) = \frac{1}{2}\) , where A > B > 0° and A + B is an acute angle, then the value of A is:

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