Question

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

• A. $\sin x$
• B. $\cos x$
• C. $0$
• D. $1$

Answer 1

Answer: (C)

Given, $f(n)=\cos n x(\sec x)^{n}$
$g(n) \sin n x(\sec x)^{n}$
$f(2020)=\cos (2020 x)(\sec x)^{2020}$
$f(2019)=\cos (2019 x)(\sec x)^{2019}$
$g(2019)=\sin (2019 x)(\sec x)^{2019}$
$f(2020)-f(2019)+\tan x g(2019)$
$=\cos (2020 x)(\sec x)^{2020}-\cos (2019 x)(\sec x)^{2019}+\frac{\sin x}{\cos x}(\sin (2019 x))(\sec x)^{2019}$
$=\cos (2020 x)(\sec x)^{2020}-\frac{(\sec x)^{2019}}{\cos x}$
$(\cos x \cos 2019 x-\sin x \sin 2019 x)$
$=\cos (2020 x)(\sec x)^{2020}-(\sec x)^{2020} \cos (2020 x)$
$=0$