## Trigonometry Tricks for Competitive Exams: Problem-solving in a Few Steps

Master trigonometry tricks and solve identities quickly! Valuable for competitive exams, SSC CGL, CHSL, CPO, and Class 10.

In competitive exams like SSC CGL, SSC CHSL, and SSC CPO, where time is the most crucial factor for success, using trigonometry tricks can significantly improve your problem-solving speed and accuracy.

Instead of relying on long, conventional methods, students should utilize their intelligence and problem-solving skills to find efficient solutions.

This is especially crucial for trigonometric identities, a common topic in Class 10 trigonometry and competitive exams.

Traditional methods often involve expanding expressions and simplifying numerous terms, which can be time-consuming and error-prone.

A more intelligent approach focuses on analyzing the problem and identifying key information to reach the solution quickly.

Let's illustrate this with a problem example, demonstrating how trigonometry shortcuts can help you solve problems efficiently.

### Trigonometric Identities: Efficient Solutions for SSC Exams

#### Problem Example

Prove the identity:

$(cos θ – cosec θ)^2 + (sin θ – sec θ)^2 = (1 – sec θ cosec θ)^2$

Try solving this problem yourself before proceeding!

You might be surprised how quickly you can solve it with a more efficient approach.

### Efficient Solution Using Trigonometry Tricks

The key to efficient problem-solving is to analyze the problem statement effectively.

This is crucial for success in competitive exams and various work environments.

#### Class 10 Trigonometry and Problem Analysis

Begin by comparing the right-hand side (RHS) expression to the left-hand side (LHS) expression, looking for similarities.

This is a direct application of the inventiveEnd State Analysis technique, where you analyze the desired end state or goal along with the starting state to guide you quickly to the solution.

#### Key Information Discovery

Our goal is to transform each expression on the LHS to match the RHS expression.

Let's start with the first expression, $(cos θ – cosec θ)^2$.

We find, if we factor $cos θ$ out of the brackets and multiply the second term by its inverse, we directly get the RHS expression as the second factor of the first expression. $cos θ$ transforms to $cos^2 θ$.

The first expression is transformed to:

$(cos θ – cosec θ)^2 = cos^2 θ(1 - sec θ cosec θ)^2$

We expect a similar result for the second expression by taking $sin θ$ out of the brackets, and that's exactly what happens:

$(sin θ – sec θ)^2 = sin^2 θ(1 - sec θ cosec θ)^2$

Now the solution is just a step away. The three steps of the efficient solution:

$(cos θ – cosec θ)^2 + (sin θ – sec θ)^2$

$= cos^2 θ (1 – sec θ cosec θ)^2 + sin^2 θ(1 – sec θ cosec θ)^2$

$= (sin^2 θ + cos^2 θ) (1 – sec θ cosec θ)^2$

$= (1 – sec θ cosec θ)^2$

This approach, based on analyzing and utilizing the similarity between the final and initial states, provides the quickest path to the solution in just three steps.

#### Reasoning behind this inventive approach: Deductive Reasoning

By comparing the RHS and LHS, we observe that both are squares, suggesting that the RHS expression is likely hidden within each of the squared expressions on the LHS.

Then we compare the RHS expression with each term on the LHS.

Taking the first term and comparing it with the first term of the RHS (which is 1), we are led to transform $cos θ$ to 1 by factoring it out of the brackets.

This deductive reasoning helps us uncover the efficient solution path.

### Conventional Solution: Why it's Less Efficient

A conventional solution usually involves expanding the square expressions, accumulating terms for simplification, and then re-transforming them back to the desired RHS expression.

This method is often lengthy and increases the likelihood of errors.

Here's how the conventional solution works for this problem:

Taking the LHS expression $(cos θ – cosec θ)^2 + (sin θ – sec θ)^2$ and expanding, we get:

$(cos θ – cosec θ)^2 + (sin θ – sec θ)^2$

$= (cos^2 θ - 2cos θ cosec θ + cosec^2 θ) + (sin^2 θ - 2sin θ sec θ + sec^2 θ)$

$= (sin^2 θ + cos^2 θ) - 2(cos θ cosec θ + sin θ sec θ) + (cosec^2 θ + sec^2 θ)$

$= 1 - 2(cos θ/sin θ + sin θ/cos θ) + (1/sin^2 θ + 1/cos^2 θ)$

$= 1 - 2(cos^2 θ + sin^2 θ)/(sin θ cos θ) + (cos^2 θ + sin^2 θ)/(sin^2 θ cos^2 θ)$

$= 1 - 2(1/(sin θ cos θ)) + (1/(sin^2 θ cos^2 θ))$

$= 1 - 2sec θ cosec θ + (sec θ cosec θ)^2$

$= (1 - sec θ cosec θ)^2$

$= RHS$

As you can see, the conventional method involves significantly more steps compared to the efficient solution using trigonometry tricks.

### Conclusion

This example demonstrates how trigonometry tricks and trigonometry shortcuts can significantly improve your problem-solving efficiency in competitive exams.

By utilizing intelligent problem-solving techniques and problem-solving techniques for quick school trigonometry solutions, you can solve trigonometric identities and other mathematical problems faster and more accurately.

Remember, always look for shorter, more efficient paths to the solution.

Practice these techniques to enhance your problem-solving skills and excel in your exams.

We have prepared inventive quick solutions for SSC CGL and other competitive exam trigonometry in the series: Quick trigonometry.

You can also access all of the 50 plus articles on quick solutions for SSC CGL math topics: Algebra, Trigonometry, Geometry and Arithmetic in Quick math.

Do use these valuable resources.