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Answer :

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.
Abidemiokingo

The simplified form of the given expression is 2tanx. Option C is correct.

Trigonometry identity

Given the following trigonometry identity:

[tex]\frac{cosx}{1-sinx} -\frac{cosx}{1+sinx} [/tex]

Simplify the expression

Find the LCM of the expression to have:

[tex]=\frac{cosx}{1-sinx} -\frac{cosx}{1+sinx} \\ =\frac{cosx(1+sinx)-(cosx(1-sinx)}{(1-sinx)(1+sinx)} \\ =\frac{cosx+cosxsinx-cosx+cosxsinx}{1-sin^2x}\\ =\frac{2cosxsins}{cos^2x } \\ =\frac{2cosxsinx}{cos^2x}\\ [/tex]

This can be simplified further to have:

[tex]=\frac{2sinxcosx}{cos^2x}\\ =\frac{2sinx}{cosx}\\ =2tanx[/tex]

Therefore the simplified form of the given expression is 2tanx.

Learn more on trigonometry function here: https://brainly.com/question/4515552

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