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Solve for [tex]x[/tex].

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

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Rewrite the following expression:

[tex]\[ \operatorname{Tan}\left(a + 5^{\circ}\right) = \sqrt{2 \sin 30^{\circ} + \sec^2 45^{\circ}} \][/tex]

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Read the lines from "The Tide Rises, The Tide Falls":

"Darkness settles on roofs and walls,
But the sea, the sea in darkness calls;"

The imagery in these lines evokes a sense of:
A. laziness
B. fear
C. mystery
D. despair


Sagot :

Sure, let's solve the given expression step-by-step:

We need to evaluate [tex]\(\operatorname{Tan}\left(a+5^{\circ}\right) \)[/tex] given that:

[tex]\[ \operatorname{Tan}\left(a+5^{\circ}\right)=\sqrt{2 \sin 30^{\circ}+\sec^2 45^{\circ}} \][/tex]

1. Calculate [tex]\(\sin 30^\circ\)[/tex]
[tex]\[ \sin 30^\circ = \frac{1}{2} = 0.5 \][/tex]

2. Calculate [tex]\(\sec 45^\circ\)[/tex]
The secant function is the reciprocal of the cosine function. Therefore:
[tex]\[ \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \][/tex]

3. Square [tex]\(\sec 45^\circ\)[/tex]
[tex]\[ \sec^2 45^\circ = (\sqrt{2})^2 = 2 \][/tex]

4. Substitute [tex]\(\sin 30^\circ\)[/tex] and [tex]\(\sec^2 45^\circ\)[/tex] into the expression inside the square root
[tex]\[ 2 \sin 30^\circ + \sec^2 45^\circ = 2 \times 0.5 + 2 = 1 + 2 = 3 \][/tex]

5. Take the square root of the expression
[tex]\[ \sqrt{3} \approx 1.73205 \][/tex]

6. Evaluate [tex]\(\operatorname{Tan}\left(a+5^{\circ}\right)\)[/tex] where the value of [tex]\(a\)[/tex] is 5
We need to find the tan of the angle [tex]\(10^\circ\)[/tex] (since [tex]\(a + 5 = 5 + 5 = 10\)[/tex]):
[tex]\[ \tan(10^\circ) \approx 0.17632698070846498 \][/tex]

So, the step-by-step solution shows that:

[tex]\[ \operatorname{Tan}\left(a+5^{\circ}\right) \approx 0.17632698070846498 \][/tex]

Therefore:

[tex]\[ \operatorname{Tan}\left(10^\circ\right) = \sqrt{2 \sin 30^{\circ} + \sec^2 45^{\circ}} \approx 0.17632698070846498 \][/tex]