If \(\frac{1-tanA}{1+tanA} = \frac{tan7° × tan11° × tan30° × tan79° × tan83°}{tan 39° × tan11° × tan60° × tan51° × tan49°}\), find cot2A
Answer & explanation
Correct answer: option 3
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