Let \(f:(-1,1) \rightarrow \mathbb{R}\) be such that \(f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}\) for \(\theta \in\left(0, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\). Then the value(s) of \(f\left(\frac{1}{3}\right)\) is (are) (A) \(1-\sqrt{\frac{3}{2}}\) (B) \(1+\sqrt{\frac{3}{2}}\) (C) \(1-\sqrt{\frac{2}{3}}\) (D) \(1+\sqrt{\frac{2}{3}}\)

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables where both sides of the equation are defined. These identities are crucial in solving various problems in trigonometry because they allow one to transform complex expressions into simpler, equivalent forms.

For instance, one fundamental identity is the Pythagorean identity, which comes from the Pythagorean theorem and states that for any angle \(\theta\): \[\sin^2\theta + \cos^2\theta = 1.\] Knowing this identity, you can derive two other useful forms: \[1 - \sin^2\theta = \cos^2\theta\] and \[1 - \cos^2\theta = \sin^2\theta.\]

Another set of identities involves reciprocal relationships, such as \[\sec\theta = \frac{1}{\cos\theta}\] and \[\csc\theta = \frac{1}{\sin\theta}.\] In the context of the JEE Advanced problem, applying the identity \(\sec(\theta) = \frac{1}{\cos(\theta)}\) to the given function \(f(\cos 4\theta)\) allows simplification of the complex expression, making it easier to find the function values.

Understanding the properties of functions is essential when working with mathematical expressions and solving equations. A function essentially relates an input to an output in a consistent way. In the textbook exercise, we're investigating a function \(f\) defined for inputs between -1 and 1.

Several key properties of functions include:

• Domain and Range: The domain refers to the set of all possible input values, while the range is the set of all potential outputs. For the function \(f\) in our case, the domain is the interval \( (-1, 1) \) and the range is \( \mathbb{R} \) (all real numbers).
• Injectivity: A function is injective (or one-to-one) if different inputs lead to different outputs.
• Surjectivity: A function is surjective (or onto) if every possible output is the image of at least one input.
• Bijectivity: A function is bijective if it is both injective and surjective, meaning there is a perfect pair for each input and output.
• Periodicity: A function is periodic if it repeats values at regular intervals of the input variable.

For JEE Advanced problems involving trigonometry, understanding these properties helps you determine the behavior of trigonometric functions and ultimately solve for unknowns.

Inverse trigonometric functions are the inverse operations of the basic trigonometric functions. They allow us to determine the angle when the value of a trigonometric function is known. The most common inverse trigonometric functions include \(\arcsin(x)\), \(\arccos(x)\), and \(\arctan(x)\).

For example, if \(\sin\theta = x\), then \(\theta = \arcsin(x)\). This is particularly useful when you have the trigonometric value but need to find the angle associated with it, which is commonly required for solving trigonometric equations. In the context of the exercise, if we know the value \(\cos 4\theta\), we would use the inverse cosine function, \(\arccos(x)\), to find the angle \(4\theta\).

Remember, the range of the inverse functions is limited to ensure they are functions (pass the vertical line test). For instance, the range of \(\arcsin(x)\) and \(\arccos(x)\) is \( [ -\frac{\pi}{2}, \frac{\pi}{2} ] \) and \( [0, \pi] \) respectively. Keeping in mind these ranges is crucial when determining the correct principle value of the angle, which is necessary for solving problems and ensuring answers fall within the correct interval.