Question

If \(0<\mathrm{A}<\pi\) and \(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A}=2\), then \(\sin ^{\mathrm{n}} \mathrm{A}+\operatorname{cosec}^{\mathrm{n}} \mathrm{A}\), where \(\mathrm{n}\) is a positive integer, is (a) \(-1\) (b) 0 (c) 1 (d) 2

Short answer

Answer: 2

Step-by-step solution

Setting Up the Equation

Since we have \(\sin A + \cosec A = 2\), we can write \(\cosec A = 2 - \sin A\). We know that \(\cosec A = \frac{1}{\sin A}\), so we can write the equation as follows: $$\frac{1}{\sin A} = 2 - \sin A$$

Solving for \(\sin A\)

To solve for \(\sin A\), we first multiply both sides of the equation by \(\sin A\): $$1 = (2 - \sin A) \cdot \sin A$$ Now, rearrange the above equation as a quadratic equation: $$\sin^2 A - 2\sin A + 1 = 0$$ This equation can be factored as: $$(\sin A - 1)^2 = 0$$ Therefore, we get \(\sin A = 1\). Note that this value is consistent with the given condition that \(0 < A < \pi\).

Substituting \(\sin A\) and Evaluating the Given Expression

Now that we found \(\sin A = 1\), we can find the value of \(\cosec A\). Since \(\cosec A = \frac{1}{\sin A}\), we have \(\cosec A = \frac{1}{1} = 1\). We are now ready to evaluate the given expression \(\sin^n A + \cosec^n A\) for any positive integer \(n\): $$\sin^n A + \cosec^n A = 1^n + 1^n = 1 + 1 = 2$$ Thus, the value of the given expression is \(\boxed{2}\), which is option (d) in the given choices.

Key concepts

Understanding Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for every value within their domains. These identities are crucial for simplifying complex expressions and solving trigonometric equations.

At the very heart of these identities is the Pythagorean identity, \[\sin^2 A + \cos^2 A = 1\], which arises from the Pythagorean theorem applied to a right-angled triangle. From this fundamental relationship, we can derive a multitude of other important identities, such as \[\cosec A = \frac{1}{\sin A}\] and its reciprocal \[\sin A = \frac{1}{\cosec A}\].

When solving problems, understanding the relationship between the sine function and its reciprocal, the cosecant function, becomes particularly useful. In our textbook exercise, we're given that \(\sin A + \cosec A = 2\), which implicitly suggests a very specific angle A for which this condition holds. Using trigonometric identities effectively allows us to manipulate the given equation to reveal this angle and subsequently solve the problem.

Mastering these identities is imperative not only for solving direct problems but also for employing them in more complex scenarios, such as when proving other mathematical claims or when dealing with integrals and derivatives in calculus.

Solving Trigonometric Equations

Trigonometric equations are mathematical statements that describe a relationship between trigonometric values that can be satisfied by some angles. Solving these equations often requires an understanding of trigonometric identities, standard trigonometric values, and algebraic manipulation skills.

In the context of our exercise, we transformed the given equation into a quadratic one, suggesting a similarity to algebraic approaches: \[\sin^2 A - 2\sin A + 1 = 0\]. Factoring this quadratic, we find that \(\sin A - 1)^2 = 0\), leading to the straightforward solution of \(\sin A = 1\).

Common Strategies for Solving Trigonometric Equations Include:

• Converting all trigonometric functions to the same function to simplify the equation.
• Applying trigonometric identities to manipulate the equation.
• Using algebraic techniques like factoring, finding common denominators, and completing the square.
• Employing graphical methods to visualize the solutions.

The value we obtained for \(\sin A\), under the condition \(0< A <\pi\), is perfectly legitimate as \(\sin A\) ranges from 0 to 1 within this interval. Once the value of \(\sin A\) is known, finding the value of \(\cosec A\), and ultimately solving the given expression for any positive integer \(n\), becomes a straightforward task.

IIT-JEE Mathematics Preparation

The Indian Institutes of Technology Joint Entrance Examination (IIT-JEE) is an annual engineering college entrance examination in India, highly competitive and challenging, especially in the field of mathematics. Preparation for this exam requires in-depth understanding and practice of various mathematical concepts, including trigonometry.

Trigonometric equations and identities are a fundamental part of the IIT-JEE mathematics syllabus. As seen in our original problem, these concepts not only test the student's understanding of basic relationships but also require creative problem-solving and algebraic manipulation skills.

Here are some tips for preparing for trigonometry in the IIT-JEE:

• Develop a strong foundation in trigonometric identities and values.
• Practice numerous problems that require the application of these identities in different ways.
• Time management during practice sessions can help in the exam when time is limited.
• Understand the graphical interpretation of trigonometric functions to gain a deeper insight.
• Regularly revise and revisit concepts to ensure they are retained until the examination.

Success in the IIT-JEE mathematics section demands not only memorizing formulas but also understanding their derivation and application in a variety of problems, very much like the exercise we discussed. This holistic approach towards learning helps with retaining concepts and developing problem-solving skills that are invaluable in cracking competitive exams such as the IIT-JEE.