If \(\frac{1-tanA}{1+tanA} = \frac{tan7° × tan11° × tan30° × tan79° × tan83°}{tan 39° × tan11° × tan60° × tan51° × tan49°}\), find cot²A

4

We know that if (A+B) = 90° then tan A × tan B = 1

ATQ,

⇒ \(\frac{1-tanA}{1+tanA} = \frac{tan7° × tan11° × tan30° × tan79° × tan83°}{tan 39° × tan11° × tan60° × tan51° × tan49°}\)

⇒ \(\frac{1-tanA}{1+tanA} = \frac{1×1×\frac{1}{\sqrt {3}}}{1×1×\sqrt {3}}\) = \(\frac{1}{3}\)

⇒ 3 - 3 tanA = 1+ tanA

⇒ 4 tanA = 2

⇒ tan A = \(\frac{1}{2}\)

Thus

⇒ cot A = \(\frac{1}{tan\;A}\) = 2

⇒ cot²A = 4