*QUESTION 6*

6.1 If $-3 \sin \beta - 2 = 0$ and $\beta \in [0^\circ;270^\circ]$, use a sketch in the correct quadrant to determine the value of: $1 + \tan^2 \beta$ without a calculator.

6.2 If, $\cos 75^\circ = m$ express each of the following in terms of $m$, showing all your working:

6.2.1 $\cos^2 105^\circ$

6.2.2 $\sin 15^\circ$

6.2.3 $\tan 15^\circ$

6.3 Given the expression:

$\frac{\cos (180^\circ - k).\sin (k - 90^\circ) - 1}{\tan^2(540^\circ + k).\sin (90^\circ + k).\cos (- k)}$

6.3.1 Simplify the expression.

6.4 Prove that:

$\frac{1+\sin \theta}{1-\sin \theta}-\frac{1-\sin \theta}{1+\sin \theta}=\frac{4\tan \theta}{\cos \theta}$

6.6 If $p=\tan A+\sin A$ and $q=\tan A-\sin A$, prove that:

$pq=\tan^2 A.\sin^2 A$

1 answer

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:

  1. \(\cos(180^\circ - k) = -\cos(k)\)
  2. \(\sin(k - 90^\circ) = -\cos(k)\)
  3. \(\tan(540^\circ + k) = \tan(k)\)
  4. \(\sin(90^\circ + k) = \cos(k)\)
  5. \(\cos(-k) = \cos(k)\)

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.