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${\sin ^2}{1^\circ } + {\sin ^2}{2^\circ } + ............ + {\sin ^2}{180^\circ } = ?$

The given expression can be written as,

$\frac{1}{2}\left[ {(1 - \cos {2^\circ }) + (1 - \cos {4^\circ }) + ................ + (1 - \cos {{360}^\circ })} \right]$

$ = \frac{1}{2}\left[ {(1 + 1 + ........ + 1) - (\cos {2^\circ } + \cos {4^\circ } + ........... + \cos {{360}^\circ })} \right]$

There are 180 terms in the series.

$ = \frac{1}{2}\left[ {180 - \frac{{\cos \left( {{2^\circ } + \frac{{180 - 1}}{2} \times {2^\circ }} \right)\sin \left( {\frac{{180 \times {2^\circ }}}{2}} \right)}}{{\sin \left( {\frac{{{2^\circ }}}{2}} \right)}}} \right]$

$ = \frac{1}{2}\left( {180 - \frac{{\cos {{181}^\circ }\sin {{180}^\circ }}}{{\sin {1^\circ }}}} \right) = 90$

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$