find the general solution of θ in 4cot2θ= cot2θ – tan2θ

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find the general solution of θ in
4cot2θ= cot²θ – tan²θ

AMRIT DASH , 10 Years ago

Grade 11

1 Answers

Lab Bhattacharjee

$\cot^2\theta-\tan^2\theta=\frac{\cos^4\theta-\sin^4\theta}{\cos^2\theta\sin^2\theta} =4\dfrac{\cos2\theta}{\sin^22\theta}=4\dfrac{\cot2\theta}{\sin2\theta} \\ \\ \text{So we have } 4\cot2\theta(\sin2\theta-1)=0 \\ \\ \text{Now if } \cot2\theta=0\implies2\theta=n\pi+\dfrac\pi2 \text{ where } n \text{ is any integer} \\ \\ \text{Otherwise, }\sin2\theta=1\implies2\theta=2m\pi+\dfrac\pi2 \text{ where } n \text{ is any integer}$ $\cot^2\theta-\tan^2\theta=\frac{\cos^4\theta-\sin^4\theta}{\cos^2\theta\sin^2\theta} =4\dfrac{\cos2\theta}{\sin^22\theta}=4\dfrac{\cot2\theta}{\sin2\theta} \\ \\ \text{So we have } 4\cot2\theta(\sin2\theta-1)=0 \\ \\ \text{Now if } \cot2\theta=0\implies2\theta=n\pi+\dfrac\pi2 \text{ where } n \text{ is any integer} \\ \\ \text{Otherwise, }\sin2\theta=1\implies2\theta=2m\pi+\dfrac\pi2 \text{ where } n \text{ is any integer} \\ \\ \text{ Clearly, the second case is a subset of the first}$