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find the general solution of θ in 4cot2 θ= cot 2 θ – tan 2 θ

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

Grade:11

1 Answers

Lab Bhattacharjee
121 Points
5 years ago
\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}
 
 

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