JEE Main Pattern Questions - Inverse Trigonometry

JEE Main Pattern Questions - Inverse Trigonometry

Hello students! Inverse trigonometry is a crucial topic for JEE Main, often appearing in problems that combine concepts from trigonometry and calculus. Mastering these problems can significantly boost your score. This lesson will guide you through common patterns, essential formulas, and effective problem-solving strategies to tackle inverse trigonometric questions with confidence.

Conceptual Explanation

Inverse trigonometric functions provide the angle whose trigonometric value you know. Understanding their properties and interrelations is key to solving complex problems. Remember the domain and range restrictions of these functions, as they're critical for determining valid solutions.

Think of $\sin^{-1}x$ as "the angle whose sine is $x$." Similarly, $\cos^{-1}x$ is "the angle whose cosine is $x$," and $\tan^{-1}x$ is "the angle whose tangent is $x$." The restrictions on the domain ensure that the inverse functions are well-defined.

Important Formulas and Derivations

1. Series involving inverse functions: $$\sum_{r=1}^{n} \tan^{-1}\frac{1}{1+r+r^2} = \tan^{-1}n$$

This formula is extremely useful for telescoping series problems. Notice that the argument of the $\tan^{-1}$ function can be rewritten.

Derivation:

We can rewrite $\frac{1}{1+r+r^2}$ as $\frac{1}{1 + r(r+1)}$. Now, try expressing it as a difference:

$$\frac{1}{1 + r(r+1)} = \frac{(r+1) - r}{1 + r(r+1)}$$

Recall the formula for $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. Then, $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy}$. So,

$$\tan^{-1}\frac{(r+1) - r}{1 + r(r+1)} = \tan^{-1}(r+1) - \tan^{-1}(r)$$

Now, substitute this into the summation:

$$\sum_{r=1}^{n} \tan^{-1}\frac{1}{1+r+r^2} = \sum_{r=1}^{n} [\tan^{-1}(r+1) - \tan^{-1}(r)]$$

This is a telescoping series, so:

$$[\tan^{-1}(2) - \tan^{-1}(1)] + [\tan^{-1}(3) - \tan^{-1}(2)] + ... + [\tan^{-1}(n+1) - \tan^{-1}(n)] = \tan^{-1}(n+1) - \tan^{-1}(1)$$

If we evaluate from $r=0$ to $n$, we have

$$\sum_{r=0}^{n} \tan^{-1}\frac{1}{1+r+r^2} = \tan^{-1}(n+1) - \tan^{-1}(0)$$

Since $\tan^{-1}(0) = 0$, and redefining $n+1$ as $n$, we have $$\sum_{r=1}^{n} \tan^{-1}\frac{1}{1+r+r^2} = \tan^{-1}n$$

2. Series involving inverse functions: $$\tan^{-1}\frac{1}{2n^2} = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$$

This formula is another trick for creating telescoping series with $\tan^{-1}$.

Derivation:

Notice that $2n^2 = 1 + (2n^2 - 1) = 1 + (2n - 1)(2n + 1)$. Therefore,

$$\tan^{-1}\frac{1}{2n^2} = \tan^{-1}\frac{1}{1 + (2n - 1)(2n + 1)}$$

We want to express this as $\tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A - B}{1 + AB}$. Thus we set

$$\tan^{-1}\frac{1}{1 + (2n - 1)(2n + 1)} = \tan^{-1}(2n + 1) - \tan^{-1}(2n - 1)$$

Then we simply substitute: $A = 2n + 1$ and $B = 2n - 1$

$$\frac{A - B}{1 + AB} = \frac{(2n + 1) - (2n - 1)}{1 + (2n + 1)(2n - 1)} = \frac{2}{1 + 4n^2 - 1} = \frac{2}{4n^2} = \frac{1}{2n^2}$$

Equations with Inverse Functions

When solving equations involving inverse trigonometric functions, isolate the inverse function and then apply the corresponding trigonometric function to both sides. Be cautious of extraneous solutions by checking your answers within the restricted domain and range.

For example, if you have $\sin^{-1}x = \theta$, then $x = \sin\theta$. Remember to consider the range of $\sin^{-1}x$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Combined Trigonometric Problems

Many JEE problems combine inverse trigonometric functions with other trigonometric functions. These often require substituting trigonometric identities to simplify the expressions. Remember that $\sin(\sin^{-1}x) = x$, $\cos(\cos^{-1}x) = x$, and $\tan(\tan^{-1}x) = x$, but only within the domain restrictions of the inverse functions.

Quick Problem-Solving Strategies

1. Recognize Patterns: Quickly identify common series or equation types.
2. Use Trigonometric Identities: Simplify expressions using identities.
3. Check Domain and Range: Validate solutions against the restrictions of inverse functions.
4. Substitute Wisely: Introduce substitutions to simplify equations, like $x = \tan\theta$.
5. Telescoping Series: If you see a summation, try to express the terms as differences.

Common Mistakes to Avoid

• Ignoring Domain and Range: Always check if your solution lies within the domain and range of the inverse trigonometric functions.
• Incorrect Simplification: Be careful while using trigonometric identities. A wrong identity can lead to an incorrect solution.
• Missing Extraneous Solutions: Substituting back into the original equation is crucial to check for extraneous solutions.
• Not Recognizing Telescoping Series: Failing to identify telescoping series in summation problems.

JEE-Specific Tricks

Trick 1: Recognize special angles like $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$ and their corresponding values in inverse trigonometric functions. This can save time during the exam.

Trick 2: Use the properties of complementary angles. For example, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ can be helpful in simplifying expressions.

Trick 3: Keep a notebook of all the important formulas and revise them regularly. This will help you recall them quickly during the exam.

By understanding these patterns, formulas, and strategies, you'll be well-equipped to tackle inverse trigonometry problems in the JEE Main exam. Keep practicing, and best of luck!