Question

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

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

Answer 1

Answer: (C)

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