Problem Solving and JEE Patterns

Mixed Problems and Revision

0:00
LearnStep 1/4

Mixed Problems and Revision

Mixed Problems and Revision: Trigonometry & Inverse Trigonometry

Welcome to the ultimate test! This lesson is all about mastering the art of tackling mixed problems involving Trigonometry and Inverse Trigonometry for the JEE Main. Why is this important? Because JEE rarely presents problems in isolation. You'll often find questions that require you to combine concepts from different modules within Trigonometry, or even mix Trigonometry with other topics like Algebra or Calculus. This lesson will equip you with the skills to identify these connections, apply the right formulas, and manage your time effectively. Get ready to level up your JEE preparation!

1. Problems Combining All Modules

JEE problems often require you to integrate concepts from multiple modules. This demands a strong understanding of the fundamental principles and the ability to recognize which concepts are relevant to a given problem. For instance, you might encounter a problem that requires you to simplify a trigonometric expression using multiple angle formulas and then solve an equation involving inverse trigonometric functions.

Example: Consider a problem where you need to find the value of sin⁡(2tan⁡−1x)\sin(2\tan^{-1}x) given that x=cos⁡(sin⁡−1y)x = \cos(\sin^{-1}y). This requires you to first understand inverse trigonometric functions, then apply the double angle formula for sine, and possibly use trigonometric identities to simplify the expression.

2. Previous Year Questions Analysis

One of the best ways to prepare for JEE is to analyze previous year's questions. This gives you insights into the types of problems asked, the difficulty level, and the areas that are frequently tested. Look for patterns in the questions and try to understand the underlying concepts being tested. This will help you prioritize your study efforts and develop effective problem-solving strategies.

Strategy: Go through past JEE Main papers focusing on Trigonometry and Inverse Trigonometry. Categorize the questions based on the concepts they test. Identify the frequently occurring concepts and formulas. Practice these types of problems extensively.

3. Common Pitfalls and How to Avoid Them

Trigonometry and Inverse Trigonometry are prone to certain common mistakes. Being aware of these pitfalls and knowing how to avoid them can significantly improve your accuracy and save you valuable time during the exam.

Pitfall: Forgetting the domain and range of inverse trigonometric functions. Remember, sin⁡−1x\sin^{-1}x is defined for x∈[−1,1]x \in [-1, 1] and its range is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. Similarly, cos⁡−1x\cos^{-1}x is defined for x∈[−1,1]x \in [-1, 1] and its range is [0,π][0, \pi], and tan⁡−1x\tan^{-1}x is defined for all real numbers and its range is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

Solution: Always check if your solution lies within the defined range and domain of the inverse trigonometric functions involved.

Pitfall: Incorrectly applying trigonometric identities. Trigonometric identities are powerful tools, but they must be applied correctly. Make sure you understand the conditions under which an identity is valid before applying it.

Solution: Double-check the identity you're using and ensure that it's applicable to the given problem. Practice applying the identities in different scenarios.

4. Time Management Strategies

Time management is crucial in JEE. You need to be able to solve problems quickly and accurately. This requires a combination of conceptual clarity, problem-solving skills, and efficient techniques.

Strategy: Practice solving problems under timed conditions. Start with a few problems and gradually increase the number of problems you solve in a given time. Identify the types of problems that take you longer to solve and focus on improving your speed in those areas. Learn to recognize problems that you can solve quickly and those that require more time. Tackle the quick problems first to build confidence and save time for the more challenging ones.

Tip: Memorize the values of trigonometric functions for common angles like 0,π6,π4,π3,π20, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}. This will save you valuable time in solving problems.

Important Formulas

Let's revise all the major formulas from the previous modules. These formulas are essential for solving mixed problems and will serve as your toolkit for tackling any Trigonometry and Inverse Trigonometry problem in JEE.

  1. Trigonometric Identities:
    • sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
    • 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta
    • 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta
  2. Compound Angle Formulas:
    • sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B
    • sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A - B) = \sin A \cos B - \cos A \sin B
    • cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B
    • cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B
    • tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
    • tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}
  3. Multiple Angle Formulas:
    • sin⁡2θ=2sin⁡θcos⁡θ=2tan⁡θ1+tan⁡2θ\sin 2\theta = 2\sin\theta \cos\theta = \frac{2\tan\theta}{1 + \tan^2\theta}
    • cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}
    • tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}
    • sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta
    • cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta
    • tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}
  4. Transformation Formulas:
    • sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C + \sin D = 2\sin\frac{C + D}{2} \cos\frac{C - D}{2}
    • sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\sin C - \sin D = 2\cos\frac{C + D}{2} \sin\frac{C - D}{2}
    • cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\cos C + \cos D = 2\cos\frac{C + D}{2} \cos\frac{C - D}{2}
    • cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C - \cos D = -2\sin\frac{C + D}{2} \sin\frac{C - D}{2}
  5. Conditional Identities:

    If A+B+C=πA + B + C = \pi, then:

    • tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A \tan B \tan C
    • cot⁡Acot⁡B+cot⁡Bcot⁡C+cot⁡Ccot⁡A=1\cot A \cot B + \cot B \cot C + \cot C \cot A = 1
    • sin⁡2A+sin⁡2B+sin⁡2C=4sin⁡Asin⁡Bsin⁡C\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C
    • cos⁡2A+cos⁡2B+cos⁡2C=−1−4cos⁡Acos⁡Bcos⁡C\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A \cos B \cos C
  6. Inverse Trigonometric Functions Formulas:
    • sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, x∈[−1,1]x \in [-1, 1]
    • tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, x∈Rx \in \mathbb{R}
    • sec⁡−1x+csc⁡−1x=π2\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}, ∣x∣≥1|x| \geq 1
    • tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right), xy < 1
    • tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy)\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right), xy > -1
    • 2tan⁡−1x=sin⁡−1(2x1+x2)2\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^2}\right), ∣x∣≤1|x| \le 1
    • 2tan⁡−1x=cos⁡−1(1−x21+x2)2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), x≥0x \ge 0
    • 2tan⁡−1x=tan⁡−1(2x1−x2)2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right), |x| < 1

Key Formula: sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

This fundamental identity relates sine and cosine. It is derived from the Pythagorean theorem on the unit circle. It is often used to simplify expressions or rewrite trigonometric functions in terms of each other.

Key Formula: tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right), xy < 1

This formula allows you to combine two inverse tangent functions into a single one. The condition xy < 1 is important because the tangent function has a period of π\pi, and this formula holds true only under this condition.

JEE-Specific Tricks

Sometimes, JEE problems can be solved more easily using specific tricks or shortcuts. These tricks are usually based on specific patterns or properties that are frequently tested in the exam.

Trick: When dealing with expressions involving sin⁡−1x+sin⁡−1y\sin^{-1}x + \sin^{-1}y or cos⁡−1x+cos⁡−1y\cos^{-1}x + \cos^{-1}y, try to convert them into tan⁡−1\tan^{-1} form using the identity tan⁡−1z=sin⁡−1(z1+z2)=cos⁡−1(11+z2)\tan^{-1}z = \sin^{-1}\left(\frac{z}{\sqrt{1+z^2}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{1+z^2}}\right). This can often simplify the problem.