Q61 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics
For $n \in \mathbf{N}$, 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$\cos x$
- Correct answerC0
- D1
Solution
Given, $f(n)=\cos n x(\sec x)^n$
$\begin{aligned} & g(n) \sin n x(\sec x)^n \\ & \quad f(2020)=\cos (2020 x)(\sec x)^{2020} \\ & f(2019)=\cos (2019 x)(\sec x)^{2019} \\ & g(2019)=\sin (2019 x)(\sec x)^{2019} \\ & f(2020)-f(2019)+\tan x g(2019) \\ & =\cos (2020 x)(\sec x)^{2020}-\cos (2019 x)(\sec x)^{2019} \\ & \quad+\frac{\sin x}{\cos x}(\sin (2019 x))(\sec x)^{2019} \\ & \quad(\cos x \cos 2019 x-\sin x \sin 2019 x) \\ & =\cos (2020 x)(\sec x)^{2020}-\frac{(\sec x)^{2019}}{\cos x} \\ & =\cos (2020 x)(\sec x)^{2020}-(\sec x)^{2020} \cos (2020 x) \\ & =0 \end{aligned}$
Hence, the correct answer is option: 'C'.