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Solve for [tex] x [/tex].

[tex] 3x = 6x - 2 [/tex]



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[tex]$\begin{aligned}(\sec A+\cos A)(\sec A-\cos A)= & \tan ^2 A \\ & +\sin ^2 A\end{aligned}$[/tex]
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Response:
[tex]\[ (\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2 A \][/tex]



Answer :

GinnyAnswer
To solve the equation:

[tex]\[ (\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2 A, \][/tex]

we will simplify each side separately and then compare the results. Here’s the step-by-step solution:

### Left Side Simplification
Consider the expression on the left side of the equation:

[tex]\[ (\sec A + \cos A)(\sec A - \cos A). \][/tex]

This expression resembles the difference of squares, which has the form [tex]\((x + y)(x - y) = x^2 - y^2\)[/tex]. Here, [tex]\( x = \sec A \)[/tex] and [tex]\( y = \cos A \)[/tex]. Applying the difference of squares formula:

[tex]\[ (\sec A + \cos A)(\sec A - \cos A) = \sec^2 A - \cos^2 A. \][/tex]

Next, recall the trigonometric identity for secant:

[tex]\[ \sec A = \frac{1}{\cos A} \implies \sec^2 A = \frac{1}{\cos^2 A}. \][/tex]

Substituting this identity in:

[tex]\[ \sec^2 A - \cos^2 A = \frac{1}{\cos^2 A} - \cos^2 A. \][/tex]

We rewrite [tex]\(\frac{1}{\cos^2 A}\)[/tex]:

[tex]\[ \frac{1}{\cos^2 A} - \cos^2 A = \frac{1 - \cos^4 A}{\cos^2 A}. \][/tex]

Notice that [tex]\(1 - \cos^4 A\)[/tex] can be factored using the difference of squares:

[tex]\[ 1 - \cos^4 A = (1 - \cos^2 A)(1 + \cos^2 A). \][/tex]

So we have:

[tex]\[ \frac{(1 - \cos^2 A)(1 + \cos^2 A)}{\cos^2 A}. \][/tex]

Using the Pythagorean identity [tex]\( \sin^2 A = 1 - \cos^2 A \)[/tex]:

[tex]\[ = \frac{\sin^2 A (1 + \cos^2 A)}{\cos^2 A}. \][/tex]

Breaking it up, we get:

[tex]\[ = \sin^2 A \cdot \frac{1 + \cos^2 A}{\cos^2 A} = \sin^2 A \left( \frac{1}{\cos^2 A} + 1 \right). \][/tex]

Remembering that [tex]\(\frac{1}{\cos^2 A} = \sec^2 A\)[/tex]:

[tex]\[ = \sin^2 A (\sec^2 A + 1). \][/tex]

### Right Side Simplification
Now consider the right side of the equation:

[tex]\[ \tan^2 A + \sin^2 A. \][/tex]

Recall that [tex]\(\tan A = \frac{\sin A}{\cos A}\)[/tex], so:

[tex]\[ \tan^2 A = \left(\frac{\sin A}{\cos A}\right)^2 = \frac{\sin^2 A}{\cos^2 A}. \][/tex]

So the right side can be written as:

[tex]\[ \tan^2 A + \sin^2 A = \frac{\sin^2 A}{\cos^2 A} + \sin^2 A = \sin^2 A \left( \frac{1}{\cos^2 A} + 1 \right). \][/tex]

Using [tex]\(\frac{1}{\cos^2 A} = \sec^2 A\)[/tex] again:

[tex]\[ = \sin^2 A (\sec^2 A + 1). \][/tex]

### Conclusion
Both sides of the original equation simplify to the same expression:

[tex]\[ (\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2 A = \sin^2 A (\sec^2 A + 1). \][/tex]

Therefore, the equality holds true, confirming that:

[tex]\[ (\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2 A. \][/tex]

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