tan alpha + 2tan 2alpha + 4tan 4alpha + 8cot | vedclass

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Question

(c) $	an \alpha + 2	an \,\,2\alpha + 4	an \,\,4\alpha + 8\cot \,8\alpha $

$ = 	an \alpha + 2	an \,2\alpha + 4\left[ {\frac{{\sin 4\alpha }}{{\

Vedclass · Accepted Answer

$\cot \,\alpha $

Vedclass · Answer

(c) $	an \alpha + 2	an \,\,2\alpha + 4	an \,\,4\alpha + 8\cot \,8\alpha $

$ = 	an \alpha + 2	an \,2\alpha + 4\left[ {\frac{{\sin 4\alpha }}{{\cos 4\alpha }} + 2\frac{{\cos \,8\alpha }}{{\sin \,8\alpha }}} ight]$

$ = 	an \alpha + 2	an 2\alpha + $

$4\left[ {\frac{{\cos \,4\alpha \,\cos \,8\alpha + \sin \,4\alpha \,\sin \,8\alpha + \cos \,4\alpha \cos \,8\alpha }}{{\sin \,8\alpha \,\cos \,4\alpha }}} ight]$

$ = 	an \,\alpha + 2	an \,2\alpha + 4\left[ {\frac{{\cos \,4\alpha + \cos \,4\alpha \,\cos \,8\alpha }}{{\sin \,8\alpha \cos \,4\alpha }}} ight]$

$ = 	an \,\alpha + 2\,	an \,2\alpha + 4\,\left[ {\frac{{\cos \,\,4a(1 + \cos \,8\alpha )}}{{\cos \,4\alpha \sin \,8\alpha }}} ight]$

$ = 	an \alpha + 2	an \,2\alpha + 4\left[ {\frac{{2{{\cos }^2}4\alpha }}{{2\sin \,4\alpha \,\,\cos \,\,4\alpha }}} ight]$

$ = 	an \,\alpha + 2	an \,2\alpha + 4\cot \,4\alpha $$ = 	an \alpha + 2(	an 2\alpha + 2\cot 4\alpha )$

$ = 	an \,\alpha + 2\left[ {\frac{{\sin \,\,2\alpha }}{{\cos 2\alpha }} + 2\frac{{\cos \,4\alpha }}{{\sin \,4\alpha }}} ight]$

$ = 	an \,\alpha + 2\left[ {\frac{{\cos \,2\alpha (1 + \cos \,4\alpha )}}{{\sin \,4\alpha \cos \,2\alpha }}} ight]$

$ = 	an \alpha + 2\cot 2\alpha = \frac{{\sin \,\alpha }}{{\cos \,\alpha }} + \frac{{2\cos \,2\alpha }}{{\sin \,2\alpha }}$

$ = \frac{{\cos \,\alpha + \cos \alpha \cos \,2\alpha }}{{\sin \,2\alpha \cos \alpha }}$

$ = \frac{{1 + \cos \,2\alpha }}{{\sin \,2\alpha }} = \frac{{2{{\cos }^2}\alpha }}{{2\sin \alpha \cos \alpha }} = \cot \,\alpha $.

$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $

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