This question asks us to find the value of the trigonometric expression cosec 15° sec 15°. To solve this, we can use fundamental trigonometric identities and the double angle formula for sine.

Understanding the Expression cosec 15° sec 15°

The expression involves the cosecant and secant functions evaluated at 15°. Let's recall the definitions of these functions in terms of sine and cosine:

• cosecant (cosec) is the reciprocal of sine: $\text{cosec } \theta = \frac{1}{\text{sin } \theta}$
• secant (sec) is the reciprocal of cosine: $\text{sec } \theta = \frac{1}{\text{cos } \theta}$

Using these definitions, we can rewrite the given expression:

$\text{cosec } 15^{\circ} \text{ sec } 15^{\circ} = \frac{1}{\text{sin } 15^{\circ}} \times \frac{1}{\text{cos } 15^{\circ}} = \frac{1}{\text{sin } 15^{\circ} \text{ cos } 15^{\circ}}$

Applying Trigonometric Identities for cosec 15° sec 15°

Now, we have the term $\text{sin } 15^{\circ} \text{ cos } 15^{\circ}$ in the denominator. This form is related to the double angle identity for sine, which is:

$\text{sin } 2\theta = 2 \text{ sin } \theta \text{ cos } \theta$

We can rearrange this identity to express $\text{sin } \theta \text{ cos } \theta$:

$\text{sin } \theta \text{ cos } \theta = \frac{1}{2} \text{ sin } 2\theta$

Let's apply this identity with $\theta = 15^{\circ}$.

$\text{sin } 15^{\circ} \text{ cos } 15^{\circ} = \frac{1}{2} \text{ sin } (2 \times 15^{\circ}) = \frac{1}{2} \text{ sin } 30^{\circ}$

Evaluating sin 30° and Calculating the Final Value

We know the standard value of $\text{sin } 30^{\circ}$.

$\text{sin } 30^{\circ} = \frac{1}{2}$

Substitute this value back into the expression for $\text{sin } 15^{\circ} \text{ cos } 15^{\circ}$:

$\text{sin } 15^{\circ} \text{ cos } 15^{\circ} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Now substitute this back into the original expression $\frac{1}{\text{sin } 15^{\circ} \text{ cos } 15^{\circ}}$:

$\text{cosec } 15^{\circ} \text{ sec } 15^{\circ} = \frac{1}{\frac{1}{4}}$

To divide by a fraction, we multiply by its reciprocal:

$\frac{1}{\frac{1}{4}} = 1 \times \frac{4}{1} = 4$

Therefore, the value of cosec 15° sec 15° is 4.

Revision Table: Key Trigonometric Concepts

Additional Information: Values for 15° and 75°

While we didn't need the individual values of sin 15° or cos 15° for this particular problem thanks to the double angle identity, it's useful to know them for other trigonometry problems. These values can be derived using sum/difference identities (e.g., $15^{\circ} = 45^{\circ} - 30^{\circ}$).

• $\text{sin } 15^{\circ} = \text{sin } (45^{\circ} - 30^{\circ}) = \text{sin } 45^{\circ} \text{ cos } 30^{\circ} - \text{cos } 45^{\circ} \text{ sin } 30^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$
• $\text{cos } 15^{\circ} = \text{cos } (45^{\circ} - 30^{\circ}) = \text{cos } 45^{\circ} \text{ cos } 30^{\circ} + \text{sin } 45^{\circ} \text{ sin } 30^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

You can verify that $\text{sin } 15^{\circ} \text{ cos } 15^{\circ} = \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = \frac{(\sqrt{6})^2 - (\sqrt{2})^2}{16} = \frac{6 - 2}{16} = \frac{4}{16} = \frac{1}{4}$, which matches our result from using the double angle identity directly. This confirms the intermediate step in our solution.