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Are these solution correct

I'll solve each problem step by step.

6.1

Step 1: Solve for sin β
Given -3 sin β - 2 = 0, we can solve for sin β: -3 sin β = 2, sin β = -2/3.

Step 2: Determine the quadrant of β
Since β ∈ [0°;270°] and sin β is negative, β must be in the third or fourth quadrant. However, since sin β = -2/3, which is negative, β is in the third or fourth quadrant. We will use this information to determine the correct quadrant.

Step 3: Draw a sketch in the correct quadrant
Drawing a sketch in the third quadrant, we can label the opposite side as -2 and the hypotenuse as 3. Using the Pythagorean theorem, we can find the adjacent side: adjacent side = √(3^2 - (-2)^2) = √(9 - 4) = √5.

Step 4: Calculate tan β
Now we can calculate tan β: tan β = opposite side / adjacent side = -2 / √5.

Step 5: Calculate 1 + tan^2 β
Using the identity 1 + tan^2 β = sec^2 β, we can calculate 1 + tan^2 β: 1 + tan^2 β = 1 + (-2/√5)^2 = 1 + 4/5 = 9/5.

The final answer is: $\boxed{\frac{9}{5}}$

6.2.1

Step 1: Express cos^2 105° in terms of m
Given cos 75° = m, we can express cos^2 105° in terms of m: cos^2 105° = cos^2 (180° - 75°) = (-cos 75°)^2 = m^2.

The final answer is: $\boxed{m^2}$

6.2.2

Step 1: Express sin 15° in terms of m
Given cos 75° = m, we can express sin 15° in terms of m: sin 15° = sin (90° - 75°) = cos 75° = m.

The final answer is: $\boxed{m}$

6.2.3

Step 1: Express tan 15° in terms of m
Given cos 75° = m, we can express tan 15° in terms of m: tan 15° = tan (90° - 75°) = cot 75° = 1 / tan 75°. Since tan 75° = sin 75° / cos 75° and sin^2 75° + cos^2 75° = 1, we can find sin 75°: sin 75° = √(1 - cos^2 75°) = √(1 - m^2). Now we can calculate tan 75°: tan 75° = √(1 - m^2) / m. Finally, we can calculate tan 15°: tan 15° = m / √(1 - m^2).

The final answer is: $\boxed{\frac{m}{\sqrt{1-m^2}}}$

6.3.1

Step 1: Simplify the expression
Given the expression: (cos(180° - k) * sin(k - 90°) - 1) / (tan^2(540° + k) * sin(90° + k) * cos(-k)), we can simplify it using trigonometric identities: cos(180° - k) = -cos k, sin(k - 90°) = -cos k, tan^2(540° + k) = tan^2 k, sin(90° + k) = cos k, cos(-k) = cos k. Now we can substitute these values into the expression: ((-cos k) * (-cos k) - 1) / (tan^2 k * cos k * cos k) = (cos^2 k - 1) / (tan^2 k * cos^2 k) = (-sin^2 k) / (tan^2 k * cos^2 k) = (-sin^2 k) / (sin^2 k / cos^2 k * cos^2 k) = -1.

The final answer is: $\boxed{-1}$

6.4

Step 1: Prove the identity
Given the identity: (1 + sin θ) / (1 - sin θ) - (1 - sin θ) / (1 + sin θ) = 4 tan θ / cos θ, we can prove it by simplifying the left-hand side: ((1 + sin θ)^2 - (1 - sin θ)^2) / ((1 - sin θ)(1 + sin θ)) = (1 + 2 sin θ +

1 answer

It looks like you've provided a detailed solution to multiple trigonometric problems. Let's break down each section to verify their correctness.

6.1

  1. Solve for sin β:

    • The equation is given as \(-3 \sin \beta - 2 = 0\).
    • Solving yields \(\sin \beta = -\frac{2}{3}\). This part is correct.

  2. Determine the quadrant of β:

    • Since \(\sin \beta\) is negative, β must be in the third or fourth quadrant. Correct analysis.

  3. Draw a sketch in the correct quadrant:

    • You label the opposite side as \(-2\) and the hypotenuse as \(3\). The adjacent side calculated as \(\sqrt{5}\) using the Pythagorean theorem is also correct.

  4. Calculate tan β:

    • You stated \(\tan \beta = \frac{-2}{\sqrt{5}}\). This is correct.

  5. Calculate \(1 + \tan^2 \beta\):

    • You calculated \(1 + \tan^2 \beta = 1 + \left(\frac{-2}{\sqrt{5}}\right)^2 = 1 + \frac{4}{5} = \frac{9}{5}\). This calculation is correct.

The final answer is indeed \(\boxed{\frac{9}{5}}\).

6.2.1

  1. Express \( \cos^2 105^\circ \) in terms of m:
    • You correctly state that \(\cos(105^\circ) = \cos(180^\circ - 75^\circ) = -\cos(75^\circ)\), and thus \(\cos^2(105^\circ) = m^2\). Your conclusion that the final answer is \(\boxed{m^2}\) is correct.

6.2.2

  1. Express \( \sin 15^\circ \) in terms of m:
    • You correctly identify \(\sin(15^\circ) = \cos(75^\circ) = m\). The answer \(\boxed{m}\) is correct.

6.2.3

  1. Express \( \tan 15^\circ \) in terms of m:
    • You use the relationship \( \tan(15^\circ) = \cot(75^\circ) = \frac{1}{\tan(75^\circ)}\).
    • You derive \( \tan(75^\circ) = \frac{\sin(75^\circ)}{\cos(75^\circ)} = \frac{\sqrt{1 - m^2}}{m}\).
    • Your final expression for \( \tan(15^\circ) = \frac{m}{\sqrt{1 - m^2}} \) is correct.

The final answer is \(\boxed{\frac{m}{\sqrt{1 - m^2}}}\).

6.3.1

  1. Simplify the expression:
    • Your steps to simplify the expression using trigonometric identities are well done. You systematically replace and simplify by using \( \cos(180° - k) = -\cos k \), \( \sin(k - 90°) = -\cos k \), etc.
    • Your final simplification leads to \( -1 \), which is correct.

The final answer is \(\boxed{-1}\).

6.4

  • You started the process of proving the identity step by step. However, the written expression ends abruptly, and I can’t see the full derivation.

Yet, given that you've expressed a solid approach by simplifying the left-hand side, you're on the right track. Typically, such identities become evident as you work through simplifications.

To summarize, the solutions provided for 6.1, 6.2.1, 6.2.2, 6.2.3, and 6.3.1 are correct. I would recommend completing step 6.4 in full to ensure clarity and correctness there as well.