MCQ
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
- A$\tan \alpha $
- B$\tan 2\alpha $
- ✓$\cot \,\alpha $
- D$\cot \,2\alpha $
$ = \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 $.
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Match each entry in List-$I$ to the correct entries in List-$II$.
| List-$I$ | List-$II$ |
| ($P$) $|z|^2$ is equal to | ($1$) $12$ |
| ($Q$) $|z-\bar{z}|^2$ is equal to | ($2$) $4$ |
| ($R$) $|z|^2+|z+\bar{z}|^2$ is equal to | ($3$) $8$ |
| ($S$) $|z+1|^2$ is equal to | ($4$) $10$ |
| ($5$) $7$ |
The correct option is: