STD 11 - 3. trignometrical ratios,functions and identities — JEE कक्षा 11 साइन्स — Question

$ = \tan \alpha + 2\tan \,2\alpha + 4\left[ {\frac{{\sin 4\alpha }}{{\cos 4\alpha }} + 2\frac{{\cos \,8\alpha }}{{\sin \,8\alpha }}} \right]$

$ = \tan \alpha + 2\tan 2\alpha + $

$4\left[ {\frac{{\cos \,4\alpha \,\cos \,8\alpha + \sin \,4\alpha \,\sin \,8\alpha + \cos \,4\alpha \cos \,8\alpha }}{{\sin \,8\alpha \,\cos \,4\alpha }}} \right]$

$ = \tan \,\alpha + 2\tan \,2\alpha + 4\left[ {\frac{{\cos \,4\alpha + \cos \,4\alpha \,\cos \,8\alpha }}{{\sin \,8\alpha \cos \,4\alpha }}} \right]$

$ = \tan \,\alpha + 2\,\tan \,2\alpha + 4\,\left[ {\frac{{\cos \,\,4a(1 + \cos \,8\alpha )}}{{\cos \,4\alpha \sin \,8\alpha }}} \right]$

$ = \tan \alpha + 2\tan \,2\alpha + 4\left[ {\frac{{2{{\cos }^2}4\alpha }}{{2\sin \,4\alpha \,\,\cos \,\,4\alpha }}} \right]$

$ = \tan \,\alpha + 2\tan \,2\alpha + 4\cot \,4\alpha $$ = \tan \alpha + 2(\tan 2\alpha + 2\cot 4\alpha )$

$ = \tan \,\alpha + 2\left[ {\frac{{\sin \,\,2\alpha }}{{\cos 2\alpha }} + 2\frac{{\cos \,4\alpha }}{{\sin \,4\alpha }}} \right]$

$ = \tan \,\alpha + 2\left[ {\frac{{\cos \,2\alpha (1 + \cos \,4\alpha )}}{{\sin \,4\alpha \cos \,2\alpha }}} \right]$

$ = \tan \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 $.