Different ways gives different results - solving tan⁡2a=3 Case 1) tan⁡2a=3=tan⁡(π3) 2a=nπ+π3 a=nπ2+π6, n∈Z Case 2) 2tan⁡a1−tan2a=3 Solving above equation gives tan⁡a=13 or −3 Thus either a=nπ+π6, n∈Z Or a=mπ−π3, m∈Z Or either I did something wrong or both these results are same.

nπ2+π6=mπ+π6⇔n=2m nπ2+π6=mπ−π3⇔3n+1=6m−2⇔n=2m−1

"Different ways gives different results - solving tan⁡2a=3..." was answered by our experts

Kaydence Huff

2022-01-24

Different ways gives different results - solving

\tan 2 a = \sqrt{3}

Case 1)

\tan 2 a = \sqrt{3} = \tan (\frac{π}{3})

2 a = n π + \frac{π}{3}

a = \frac{n π}{2} + \frac{π}{6}, n \in Z

Case 2)

\frac{2 \tan a}{1 - \tan^{2} a} = \sqrt{3}

Solving above equation gives

\tan a = \frac{1}{\sqrt{3}} or - \sqrt{3}

Thus either

a = n π + \frac{π}{6}, n \in Z

a = m π - \frac{π}{3}, m \in Z

Or either I did something wrong or both these results are same.

caoireoilns

Beginner2022-01-25Added 12 answers

\frac{n π}{2} + \frac{π}{6} = m π + \frac{π}{6} \Leftrightarrow n = 2 m

\frac{n π}{2} + \frac{π}{6} = m π - \frac{π}{3} \Leftrightarrow 3 n + 1 = 6 m - 2 \Leftrightarrow n = 2 m - 1

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?