For n in N,if f(n) = (cos nx)(sec x)^n and g( | vedclass

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(C) Given,$f(n) = \cos(nx)(\sec x)^n$ and $g(n) = \sin(nx)(\sec x)^n$. We need to evaluate $f(2020) - f(2019) + 	an x \cdot g(2019)$. Substitute the

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(C) Given,$f(n) = \cos(nx)(\sec x)^n$ and $g(n) = \sin(nx)(\sec x)^n$. We need to evaluate $f(2020) - f(2019) + 	an x \cdot g(2019)$. Substitute the expressions: $f(2020) - f(2019) + 	an x \cdot g(2019) = \cos(2020x)(\sec x)^{2020} - \cos(2019x)(\sec x)^{2019} + \frac{\sin x}{\cos x} \cdot \sin(2019x)(\sec x)^{2019}$. Factor out $(\sec x)^{2019}$ from the last two terms: $= \cos(2020x)(\sec x)^{2020} - (\sec x)^{2019} \left[ \cos(2019x) - \frac{\sin x \sin(2019x)}{\cos x} ight]$. Simplify the term inside the bracket: $= \cos(2020x)(\sec x)^{2020} - (\sec x)^{2019} \left[ \frac{\cos x \cos(2019x) - \sin x \sin(2019x)}{\cos x} ight]$. Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $= \cos(2020x)(\sec x)^{2020} - (\sec x)^{2019} \left[ \frac{\cos(2019x + x)}{\cos x} ight]$. $= \cos(2020x)(\sec x)^{2020} - (\sec x)^{2019} \left[ \frac{\cos(2020x)}{\cos x} ight]$. Since $\frac{1}{\cos x} = \sec x$: $= \cos(2020x)(\sec x)^{2020} - \cos(2020x)(\sec x)^{2019} \cdot \sec x$. $= \cos(2020x)(\sec x)^{2020} - \cos(2020x)(\sec x)^{2020} = 0$.

For $n \in N$,if $f(n) = (\cos nx)(\sec x)^n$ and $g(n) = (\sin nx)(\sec x)^n$,then $f(2020) - f(2019) + (\tan x)g(2019) =$

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