Let's address the questions one by one.
6.1
To solve the equation \(-3 \sin \beta - 2 = 0\):
\[ -3 \sin \beta = 2 \quad \Rightarrow \quad \sin \beta = -\frac{2}{3} \]
Since \(\beta \in [0^\circ; 270^\circ]\), we know that \(\sin \beta\) is negative in the third quadrant. Therefore, \(\beta\) must be in the third quadrant.
Now, to find \(1 + \tan^2 \beta\), we can use the identity:
\[ 1 + \tan^2 \beta = \sec^2 \beta \]
First, we need to calculate \(\cos \beta\) using the Pythagorean identity:
\[ \sin^2 \beta + \cos^2 \beta = 1 \]
Substituting \(\sin \beta\):
\[ \left(-\frac{2}{3}\right)^2 + \cos^2 \beta = 1 \] \[ \frac{4}{9} + \cos^2 \beta = 1 \quad \Rightarrow \quad \cos^2 \beta = 1 - \frac{4}{9} = \frac{5}{9} \]
Now, since \(\beta\) is in the third quadrant, \(\cos \beta\) is negative:
\[ \cos \beta = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3} \]
Now we can find \(\sec^2 \beta\):
\[ \sec^2 \beta = \frac{1}{\cos^2 \beta} = \frac{1}{\frac{5}{9}} = \frac{9}{5} \]
Thus,
\[ 1 + \tan^2 \beta = \sec^2 \beta = \frac{9}{5} \]
6.2
Given \(\cos 75^\circ = m\):
6.2.1 \(\cos^2 105^\circ\)
Using the identity \(\cos(105^\circ) = \cos(90^\circ + 15^\circ) = -\sin 15^\circ\):
Now, we can express \(\sin 15^\circ\) in terms of \(\cos 75^\circ\).
Using the co-function identity:
\[ \sin 15^\circ = \cos(90^\circ - 15^\circ) = \cos 75^\circ = m \]
Thus,
\(\cos^2 105^\circ = \sin^2 15^\circ = m^2\).
6.2.2 \(\sin 15^\circ\)
Using the sine and cosine relationship from above:
\[ \sin 15^\circ = \cos 75^\circ = m \]
6.2.3 \(\tan 15^\circ\)
Using the identity \(\tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ}\):
We know: \(\cos 15^\circ = \sin(75^\circ) = \sin(90^\circ - 15^\circ) = \cos 75^\circ = m\).
To find \(\sin 15^\circ\) already established as \(m\):
Thus,
\[ \tan 15^\circ = \frac{m}{m} = 1 \]
6.3
We want to simplify:
\[ \frac{\cos(180^\circ - k)\sin(k - 90^\circ) - 1}{\tan^2(540^\circ + k)\sin(90^\circ + k)\cos(-k)} \]
Using trigonometric identities:
- \(\cos(180^\circ - k) = -\cos(k)\)
- \(\sin(k - 90^\circ) = -\cos(k)\)
- \(\tan(540^\circ + k) = \tan(k)\)
- \(\sin(90^\circ + k) = \cos(k)\)
- \(\cos(-k) = \cos(k)\)
Substituting these in:
\[ = \frac{-\cos(k)(-\cos(k)) -1}{\tan^2(k) \cos(k) \cos(k)} = \frac{\cos^2(k) - 1}{\tan^2(k) \cos^2(k)} \]
Using \(\cos^2(k) - 1 = -\sin^2(k)\):
\[ = \frac{-\sin^2(k)}{\frac{\sin^2(k)}{\cos^2(k)} \cos^2(k)} = -1 \]
6.4
To prove:
\[ \frac{1+\sin \theta}{1-\sin \theta}-\frac{1-\sin \theta}{1+\sin \theta}=\frac{4\tan \theta}{\cos \theta} \]
Find a common denominator (product of the two binomials):
\[ = \frac{(1+\sin \theta)^2 - (1-\sin \theta)^2}{(1-\sin\theta)(1+\sin\theta)} \]
Using the difference of squares:
\[ = \frac{(1+\sin \theta + (1-\sin \theta))(1+\sin\theta - (1-\sin\theta))}{(1-\sin \theta)(1+\sin \theta)} = \frac{2 \sin \theta}{\cos^2 \theta} \]
So:
\[ = \frac{2 \sin \theta}{\cos^2 \theta} = \frac{4\tan \theta}{\cos \theta} \]
6.6
To show that \(pq = \tan^2 A \sin^2 A\):
\[ p = \tan A + \sin A, \quad q = \tan A - \sin A \] \[ pq = (\tan A + \sin A)(\tan A - \sin A) = \tan^2 A - \sin^2 A \]
Using \(\tan^2 A = \frac{\sin^2 A}{\cos^2 A}\):
Replace \(\tan^2 A - \sin^2 A\):
\[ = \frac{\sin^2 A}{\cos^2 A} - \sin^2 A = \frac{\sin^2 A - \sin^2 A\cos^2 A}{\cos^2 A} \] Factoring out \(\sin^2 A\):
\[ = \frac{\sin^2 A(1 - \cos^2 A)}{\cos^2 A} = \frac{\sin^2 A \sin^2 A}{\cos^2 A} = \tan^2 A \sin^2 A \]
Each of these problems has been tackled step by step, displaying necessary calculations and trigonometric identities.