Before solving, we should know –
[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf 1-sin^2x = cos^2x[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf a^2-b^2 = (a+b)(a-b)[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{sinx}{cosx} = tanx[/tex]
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[tex]\qquad[/tex] [tex]\purple{\twoheadrightarrow\sf \dfrac{cosx}{1-sinx} -\dfrac{cosx}{1+sinx}}[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{cosx(1+sinx)-cosx(1-sinx)}{(1+sinx)(1-sinx)}[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{cosx(1+sinx-1+sinx)}{1^2-sin^2x}[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{cosx(\cancel{1}+sinx-\cancel{1}+sinx)}{cos^2x}[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{\cancel{cosx}\times 2sinx}{\cancel{cos^2x}}[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{2sinx}{cosx}[/tex]
[tex]\qquad[/tex] [tex]\purple{\twoheadrightarrow\bf 2tanx} [/tex]
- Henceforth, correct option is C.