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### Prove that, ${\left( {\frac{{1 + \sin \theta + i\cos \theta }}{{1 + \sin \theta - i\cos \theta }}} \right)^n} =$ $\cos n\left( {\frac{\pi }{2} - \theta } \right) + i\sin n\left( {\frac{\pi }{2} - \theta } \right)$

LHS = ${\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{\{ 1 + (\sin \theta - i\cos \theta )\} (\sin \theta + i\cos \theta )}}} \right]^n}$

$= {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(\sin \theta + i\cos \theta ) + \{ {{\sin }^2}\theta - {{(i\cos \theta )}^2}\} }}} \right]^n}$

$= {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(\sin \theta + i\cos \theta ) + ({{\sin }^2}\theta + {{\cos }^2}\theta )}}} \right]^n}$

$= {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(1 + \sin \theta + i\cos \theta )}}} \right]^n}$

$= {(\sin \theta + i\cos \theta )^n}$

$= {\left\{ {\cos \left( {\frac{\pi }{2} - \theta } \right) + i\sin \left( {\frac{\pi }{2} - \theta } \right)} \right\}^n}$

$= \cos n\left( {\frac{\pi }{2} - \theta } \right) + i\sin n\left( {\frac{\pi }{2} - \theta } \right)$

= RHS

### $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, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$