Prove :
[tex] \large{ \tt{ \frac{1}{ \sec \alpha + 1 } - \frac{ \cos\alpha }{ { \sin^{2} \alpha } } = \frac{ \cos \alpha }{ { \sin}^{2} \alpha } - \frac{1}{ \sec \alpha - 1} }}[/tex]
*Irrelevant / Random answers will be reported!​

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

[tex]\displaystyle \frac{1}{\sec\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}[/tex]

We can convert sec(α) to 1 / cos(α):

[tex]\displaystyle \frac{1}{1/\cos\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}[/tex]

Multiply both layers of the first fraction by cos(α):

[tex]\displaystyle \frac{\cos\alpha}{1+\cos\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}[/tex]

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

[tex]\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}[/tex]

Simplify:

[tex]\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{1-\cos^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}[/tex]

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

[tex]\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{\sin^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}[/tex]

Subtract:

[tex]\displaystyle \frac{\cos\alpha(1-\cos\alpha)-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}[/tex]

Distribute:

[tex]\displaystyle \frac{\cos\alpha-\cos^2\alpha-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}[/tex]

Rewrite:

[tex]\displaystyle \frac{(\cos\alpha)-(\cos^2\alpha+\cos\alpha)}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}[/tex]

Split:

[tex]\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos^2\alpha+\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}[/tex]

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

[tex]\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{1-\cos^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}[/tex]

Factor:

[tex]\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{(1-\cos\alpha)(1+\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}[/tex]

Cancel:

[tex]\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha}{(1-\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}[/tex]

Divide the second fraction by cos(α):

[tex]\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}=\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}[/tex]

Hence proven.

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