Trigonometrical Equations
normal

The solution of $tan\,\, 2\theta\,\, tan\theta = 1$ is

A

$\frac{\pi }{3}$

B

$(6n \pm 1)\frac{\pi }{6}$

C

$(4n \pm 1)\frac{\pi }{6}$

D

$(2n \pm 1)\frac{\pi }{6}$

Solution

$\tan 2 \theta \tan \theta=1 \Rightarrow \frac{2 \tan \theta}{1-\tan ^{2} \theta} \cdot \tan \theta=1$

$\Rightarrow \quad 2 \tan ^{2} \theta=1-\tan ^{2} \theta \Rightarrow 3 \tan ^{2} \theta=1$

$\Rightarrow \quad \tan \theta=\pm \frac{1}{\sqrt{3}}=\tan \left(\pm \frac{\pi}{6}\right)$

$\Rightarrow \quad \theta=n \pi \pm \frac{\pi}{6} \quad(m \in Z)$

$=(6 n \pm 1) \frac{\pi}{6}$

$\mathrm{Or}$

$\tan 2 \theta=\cot \theta=\tan \left(\frac{\pi}{2}-\theta\right)$

$\Rightarrow \quad 2 \theta=n \pi+\frac{\pi}{2}-\theta$

$\Rightarrow \quad 3 \theta=n \pi+\frac{\pi}{2}$

$\Rightarrow \quad \theta=\frac{\mathrm{n} \pi}{3}+\frac{\pi}{6}=(2 \mathrm{n}+1) \frac{\pi}{6}$

Standard 11
Mathematics

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