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The image presents a mathematical problem set, specifically focusing on trigonometry. The content is organized into sections labeled with numbers and letters, each addressing a distinct aspect of trigonometric functions and identities.

Section 6.1:
This section begins by stating an equation involving sine: $-3 \sin \beta - 2 = 0$, with the constraint that $\beta \in [0^\circ;270^\circ]$. It then asks to use a sketch in the correct quadrant to determine the value of $1 + \tan^2 \beta$ without using a calculator.

Section 6.2:
Given $\cos 75^\circ = m$, this section requests expressions for $\cos^2 105^\circ$, $\sin 15^\circ$, and $\tan 15^\circ$ in terms of $m$, along with detailed working.

Section 6.3:
This part provides an expression involving cosine and sine functions and asks to simplify it.

Section 6.4:
A proof is required to show that $\frac{1+\sin \theta}{1-\sin \theta} - \frac{1-\sin \theta}{1+\sin \theta} = \frac{4 \tan \theta}{\cos \theta}$.

Section 6.6:
Finally, given $p = \tan A + \sin A$ and $q = \tan A - \sin A$, it asks to prove that $pq = \tan^2 A \cdot \sin^2 A$.

The image concludes with a total point value of [41] for all the questions combined

1 answer

Let's go through each section step-by-step, providing solutions and explanations for the trigonometric problems presented.

Section 6.1

Problem: Solve the equation \( -3 \sin \beta - 2 = 0 \) for \( \beta \in [0^\circ, 270^\circ] \) and find \( 1 + \tan^2 \beta \).

Solution:

  1. Solve for \( \sin \beta \): \[ -3 \sin \beta - 2 = 0 \implies -3 \sin \beta = 2 \implies \sin \beta = -\frac{2}{3} \]
  2. Since \( \sin \beta \) is negative, \( \beta \) must lie in either the third or fourth quadrant. The valid range is \( [180^\circ, 270^\circ] \).
  3. Using the identity \( 1 + \tan^2 \beta = \sec^2 \beta \): \[ \sin^2 \beta = \left(-\frac{2}{3}\right)^2 = \frac{4}{9} \]
  4. Calculate \( \cos^2 \beta \): \[ \cos^2 \beta = 1 - \sin^2 \beta = 1 - \frac{4}{9} = \frac{5}{9} \]
  5. Now, find \( \tan^2 \beta \): \[ \tan^2 \beta = \frac{\sin^2 \beta}{\cos^2 \beta} = \frac{\frac{4}{9}}{\frac{5}{9}} = \frac{4}{5} \]
  6. Finally, calculate \( 1 + \tan^2 \beta \): \[ 1 + \tan^2 \beta = 1 + \frac{4}{5} = \frac{9}{5} \]

Section 6.2

Problem: Given \( \cos 75^\circ = m \), derive expressions for \( \cos^2 105^\circ, \sin 15^\circ, \) and \( \tan 15^\circ \).

Solution:

  1. Use the identity \( \cos(90^\circ + x) = -\sin(x) \): \[ \cos 105^\circ = \cos(90^\circ + 15^\circ) = -\sin 15^\circ \]
  2. Using \( \sin^2 x + \cos^2 x = 1 \): \[ \cos^2 105^\circ = \sin^2 15^\circ \]
  3. Next, \( \sin 15^\circ \) can be found using the angle difference identity \( \sin(45^\circ - 30^\circ) \): \[ \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} \]
  4. For \( \tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} \) and using \( \cos 15^\circ = \cos(45^\circ - 30^\circ) \): \[ \cos 15^\circ = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} \]
  5. Hence, \[ \tan 15^\circ = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} \]

Section 6.3

Problem: Simplify the expression \( \cos^2 x - \sin^2 x \).

Solution: Using the identity for cosine of double angle: \[ \cos^2 x - \sin^2 x = \cos(2x) \]

Section 6.4

Problem: Prove \( \frac{1+\sin \theta}{1-\sin \theta} - \frac{1-\sin \theta}{1+\sin \theta} = \frac{4 \tan \theta}{\cos \theta} \).

Solution:

  1. Combine the fractions: \[ \frac{(1 + \sin \theta)^2 - (1 - \sin \theta)^2}{(1 - \sin \theta)(1 + \sin \theta)} \]
  2. Expand the numerator, \( (a+b)^2 - (a-b)^2 \): \[ = \sin^2 \theta + 2 \sin \theta = 2 \sin \theta \] The denominator simplifies to \( 1 - \sin^2 \theta = \cos^2 \theta \).
  3. Thus, we have: \[ \frac{2 \sin \theta}{\cos^2 \theta} = 2 \tan \theta \sec \theta = \frac{4 \tan \theta}{\cos \theta} \]

Section 6.6

Problem: Prove that \( pq = \tan^2 A \cdot \sin^2 A \).

Solution:

  1. Multiply \( p \) and \( q \): \[ pq = (\tan A + \sin A)(\tan A - \sin A) = \tan^2 A - \sin^2 A \]
  2. Use the identity \( \tan^2 A = \frac{\sin^2 A}{\cos^2 A} \): \[ pq = \frac{\sin^2 A}{\cos^2 A} - \sin^2 A = \sin^2 A \left(\frac{1 - \cos^2 A}{\cos^2 A}\right) = \sin^2 A \cdot \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \cdot \sin^2 A \]

These solutions succinctly cover each section of the problem set, demonstrating fundamental trigonometric identities and relationships.