Question

Which of these expressions is the simplified form of the expression (StartFraction sine (x) Over 1 minus cosine squared (x) EndFraction) tangent (StartFraction x Over 2 EndFraction) ?

Answer 1

Given:

$\dfrac{\sin (x)}{1-\cos ^2 (x)}\tan \left(\dfrac{x}{2}\right)$.

To find:

The expression, which is equivalent to the given expression.

Solution:

We have,

$\dfrac{\sin (x)}{1-\cos ^2 (x)}\tan \left(\dfrac{x}{2}\right)$

$=\dfrac{\sin (x)}{1^2-\cos ^2 (x)}\times \dfrac{1-\cos (x)}{\sin (x)}$         $[\because \tan \left(\dfrac{x}{2}\right)=\dfrac{1-\cos (x)}{\sin (x)}]$

$=\dfrac{\sin (x)}{(1-\cos (x))(1+\cos (x))}\times \dfrac{1-\cos (x)}{\sin (x)}$         $[\because a^2-b^2=(a-b)(a+b)]$

Cancel out the common factors.

$=\dfrac{1}{1+\cos (x)}$

Therefore, the correct option is C.

Answer 2

    Given question is incomplete; find the complete question in the attachment.

    Simplified form of the given trigonometric expression will be represented by Option C.

Given expression is,

• $[\frac{\text{sin(x)}}{1-\text{cos}^2(x)}]\tan(\frac{x}{2} )$

$[\frac{\text{sin(x)}}{1-\text{cos}^2(x)}]\tan(\frac{x}{2} )=\frac{\text{sin(x)}}{1-\text{cos}^2(x)}\times \frac{1-\text{cos}x}{\text{sinx}}$ [Since. $\text{tan}\frac{x}{2} =\frac{1-\text{cosx}}{sinx}$]

                          $=\frac{1-\text{cos}x}{(1-\text{cosx})(1+\text{cosx})}$

                          $=\frac{1}{1+\text{cosx}}$

   Therefore, Option C will be the correct option.

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