Question

Please select the best answer from the choices provided. PLZ HELP ASAP>WILL MARK BRAINLIEST

Answer 1

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

Trigonometry identity

Given the following trigonometry identity:

$\frac{cosx}{1-sinx} -\frac{cosx}{1+sinx}$

Simplify the expression

Find the LCM of the expression to have:

$=\frac{cosx}{1-sinx} -\frac{cosx}{1+sinx} \\ =\frac{cosx(1+sinx)-(cosx(1-sinx)}{(1-sinx)(1+sinx)} \\ =\frac{cosx+cosxsinx-cosx+cosxsinx}{1-sin^2x}\\ =\frac{2cosxsins}{cos^2x } \\ =\frac{2cosxsinx}{cos^2x}$

This can be simplified further to have:

$=\frac{2sinxcosx}{cos^2x}\\ =\frac{2sinx}{cosx}\\ =2tanx$

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

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

Answer 2

Before solving, we should know –

$\qquad$ $\twoheadrightarrow\sf 1-sin^2x = cos^2x$

$\qquad$ $\twoheadrightarrow\sf a^2-b^2 = (a+b)(a-b)$

$\qquad$ $\twoheadrightarrow\sf \dfrac{sinx}{cosx} = tanx$

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$\qquad$ $\purple{\twoheadrightarrow\sf \dfrac{cosx}{1-sinx} -\dfrac{cosx}{1+sinx}}$

$\qquad$ $\twoheadrightarrow\sf \dfrac{cosx(1+sinx)-cosx(1-sinx)}{(1+sinx)(1-sinx)}$

$\qquad$ $\twoheadrightarrow\sf \dfrac{cosx(1+sinx-1+sinx)}{1^2-sin^2x}$

$\qquad$ $\twoheadrightarrow\sf \dfrac{cosx(\cancel{1}+sinx-\cancel{1}+sinx)}{cos^2x}$

$\qquad$ $\twoheadrightarrow\sf \dfrac{\cancel{cosx}\times 2sinx}{\cancel{cos^2x}}$

$\qquad$ $\twoheadrightarrow\sf \dfrac{2sinx}{cosx}$

$\qquad$ $\purple{\twoheadrightarrow\bf 2tanx}$

• Henceforth, correct option is C.