## Faster Solutions for SSC CGL Trigonometry Questions: Inventive Cosine Trigo Problem-Solving Technique

Learn to solve SSC CGL cosine trigo problems faster with a unique technique. Save time on your exam by applying this efficient problem-solving approach.

Some problems may seem simple at first glance and easy to solve with standard methods. However, conventional solutions often involve unnecessary steps. Here, we present both a traditional method and a faster, inventive approach for solving SSC CGL cosine trigo problems using a powerful technique.

The key to this inventive approach lies in recognizing patterns between related trigonometric functions, known as friendly trigonometric function pairs. These pairs allow us to simplify problems quickly by reducing unnecessary steps.

### Quick Solution for SSC CGL Trigonometry Problem Example on Cosine of an Acute Angle

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

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

#### Conventional Approach for Solving SSC CGL Cosine Trigo Problems

This approach primarily uses the identity $ cosec^2 \theta - cot^2 \theta = 1 $ by squaring the given equation. The goal is to isolate $cot \theta$ and solve for $ sec \theta $ using the identity $ 1 + tan^2 \theta = sec^2 \theta $. This requires a second operation of squaring increasing the number of steps.

**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 for SSC CGL Cosine Trigo Problem Solving: Efficient Approach

Using the principle of **friendly trigonometric function pairs concepts**, we identify a faster route to the solution. Recognizing useful patterns early allows for simplification of expressions and efficient cosine trigo problem solving.

**The friendly trigonometric function pairs concepts:**

In the well-known identity $cosec^2 \theta - cot^2 \theta = 1$, the LHS can be broken up into two factors leading to a powerful relation:

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

This shows that the functions, $cosec \theta$ and $cot \theta$ **form a friendly trigonometric function pair,** highly useful in cosine trigo problem solving.* This is inventive trigonometric problem solving.*

**Given:** $ 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: Efficient Cosine Trigo Problem Solving for SSC CGL

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.

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