You are here

How to Solve an SSC CGL Trigonometry Problem Faster 7

How to Solve an SSC CGL Trigonometry Problem Faster 7

Streamlining Inventive Solutions for SSC CGL Trigonometry Problems

Can you solve the trigonometry problem in 25 seconds? We'll show you an inventive approach that skips traditional, lengthy calculations to solve it faster.

Some problems might not appear challenging at first glance and seem quick to solve using standard methods. However, these conventional solutions often involve unnecessary steps. Here, we present both a traditional method and a faster, inventive approach utilizing an embedded useful pattern and powerful problem-solving technique.


Chosen Problem

If $ cot \theta + cosec \theta = 3 $, and $ \theta $ is an acute angle, find $ cos \theta $.

  1. $1$
  2. $\displaystyle\frac{1}{2}$
  3. $\displaystyle\frac{4}{5}$
  4. $\displaystyle\frac{3}{4}$

Conventional Approach

This approach depends primarily on using the identity $ cosec^2 \theta - cot^2 \theta = 1 $ by squaring the given equation. Thus $cot \theta$ is isolated to obtain $tan \theta$. Aim is to solve for $ sec \theta $ using the identity $ 1 + tan^2 \theta = sec^2 \theta $. It needs a second operation of squaring.

Given: $ cot \theta + cosec \theta = 3 $

Thus, $ cosec \theta = 3 - cot \theta $

Squaring both sides, we get:

$ cosec^2 \theta = 9 - 6cot \theta + cot^2 \theta $

Or, $ 9 - 6cot \theta = cosec^2 \theta - cot^ \theta = 1 $

Or, $ 6cot \theta = 8 $

Or, $ tan \theta =\displaystyle\frac{3}{4} $.

Squaring again and adding 1:

$ tan^2 \theta + 1 = sec^2 \theta = \displaystyle\frac{25}{16} $

As $ \theta $ is acute, $ sec \theta $ is positive:

$ sec \theta = \displaystyle\frac{5}{4} $

Therefore, $ cos \theta = \displaystyle\frac{4}{5} $.

Answer: Option c: $ \displaystyle\frac{4}{5} $.

Inventive Solution: Efficient Problem-Solving

Using the principle of friendly trigonometric function pairs, we identify a faster route to the solution. The key is recognizing useful patterns and simplifying expressions early.

The friendly trigonometric function pair:

In the well-known identity $cosec^2 \theta - cot^2 \theta = 1$, the LHS is broken up into two factors and a powerful relation follows:

$cosec \theta + cot \theta = \displaystyle\frac{1}{cosec \theta - cot \theta}$

This way, the two functions $cosec \theta$ and $cot \theta$ form a friendly trigonometric function pair highly useful in simplifying and solving trigonometric problems elegantly and quickly. This is inventive trigonometric problem solving.

Given expression: $ cot \theta + cosec \theta = 3 $

Using the principle:

$\displaystyle\frac{1}{cosec \theta - cot \theta} = 3 $.

Thus, $ cosec \theta - cot \theta = \displaystyle\frac{1}{3} $.

Adding the two equations eliminates $ cot \theta $, simplifying to:

$ 2cosec \theta = 3 + \displaystyle\frac{1}{3} = \displaystyle\frac{10}{3} $

So, $ cosec \theta = \displaystyle\frac{5}{3} $, and therefore:

$ sin \theta = \displaystyle\frac{3}{5} $

$ cos \theta = \sqrt{1 - \left(\displaystyle\frac{3}{5}\right)^2} =\displaystyle \frac{4}{5} $, as $\theta$ is an acute angle.

Answer: Option c: $\displaystyle \frac{4}{5} $.

Solved in a minimum number of six steps using only linear first order trigonometric functions, bypassing the need of squaring twice and saving as many as four steps.

Conclusion

This refined approach not only simplifies the problem but also demonstrates the power of understanding and applying trigonometric identities effectively. By focusing on linear expressions and minimizing the order of terms, the problem-solving process is streamlined, achieving solutions more efficiently faster.


Guided help on Trigonometry in Suresolv

To get the best results out of the extensive range of articles of tutorials, questions and solutions on Trigonometry in Suresolv, follow the guide,

Reading and Practice Guide on Trigonometry in Suresolv for SSC CHSL, SSC CGL, SSC CGL Tier II and Other Competitive exams.

The guide list of articles is up-to-date.