A) \[\tan n\theta \]
B) \[\cot n\theta \]
C) \[i\tan n\theta \]
D) \[i\cot n\theta \]
Correct Answer: C
Solution :
Given, \[z=cos\text{ }\theta +i\text{ }sin\,\theta \] Then, \[{{z}^{2n}}={{(\cos \theta +i\sin \theta )}^{2n}}\] \[=\cos 2n\theta +i\sin 2n\theta \] [by Demoiver's theorem] \[\therefore \]\[\frac{{{z}^{2n}}-1}{{{z}^{2n}}+1}=\frac{\cos 2n\theta +i\sin 2n\theta -1}{\cos 2n\theta +i\sin 2n\theta +1}\] \[=\frac{1-2{{\sin }^{2}}n\theta +2\sin n\theta \cos n\theta -1}{2{{\cos }^{2}}n\theta -1+2i\sin n\theta \cos n\theta +1}\] \[=\frac{i2\sin n\theta \cos n\theta -2{{\sin }^{2}}n\theta }{2{{\cos }^{2}}n\theta +2i\sin n\theta \cos n\theta }\] \[=\frac{2\sin n\theta (i\cos n\theta -\sin n\theta )}{2\cos n\theta (\cos n\theta +i\sin n\theta )}\] \[=\frac{i\sin n\theta (\cos n\theta +i\sin n\theta )}{\cos n\theta (\cos n\theta +i\sin n\theta )}\] \[=i\tan n\theta \]
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