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Question: If \(\Rightarrow\), then the general value of \(\theta = n\pi,n \in Z\)is....

If \Rightarrow, then the general value of θ=nπ,nZ\theta = n\pi,n \in Zis.

A

sec2θ=1cos2θ=1+tan2θ1tan2θ\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1 + \tan^{2}\theta}{1 - \tan^{2}\theta}

B

tan2θ+1+tan2θ1tan2θ=1\tan^{2}\theta + \frac{1 + \tan^{2}\theta}{1 - \tan^{2}\theta} = 1

C

\Rightarrow

D

None of these

Answer

sec2θ=1cos2θ=1+tan2θ1tan2θ\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1 + \tan^{2}\theta}{1 - \tan^{2}\theta}

Explanation

Solution

\Rightarrow

tanθ(tanθ1)3(tanθ1)=0\tan\theta(\tan\theta - 1) - \sqrt{3}(\tan\theta - 1) = 0 \Rightarrow

(tanθ3)(tanθ1)=0(\tan\theta - \sqrt{3})(\tan\theta - 1) = 0 \Rightarrow.