Trigonometry is one of the most interesting chapters of Quantitative Aptitude section. Basically, it is a part of SSC syllabus. Today I will tell you the easy method to learn all the basics of trigonometry i.e. Trigonometric Ratios, facts and formulas.

asked 2021-02-09
Trigonometry is one of the most interesting chapters of Quantitative Aptitude section. Basically, it is a part of SSC syllabus. Today I will tell you the easy method to learn all the basics of trigonometry i.e. Trigonometric Ratios, facts and formulas.

Answers (1)


Trigonometric Ratios

There are six trigonometric ratios. First three are the primary functions and last three are just the reciprocals of above three. Those are written as follows:

\((\sin \theta)\)

\((\cos \theta)\)

\((\tan \theta)\)

\((\cot \theta)\)

\((\sec \theta)\)

\((\cos \sec \theta)\)

See the following image, You are already well versed with this image. This is a basic right angled triangle.
Points to remember:
1. Usually, student got confused between perpendicular and base. Always remember that out of three sides, perpendicular is that side which is opposite to the given angle.
In following figure, perpendicular (P) is the side which is opposite to the purple shaded angle(given angle)
2. Side opposite to the right angle(shaded red)  is hypotenuse (H).
3. Remaining side is known as base (B).
Now, you must be knowing that all the trigonometric ratios can be derived from these sides. Let us see how.
\(\sin \theta = \frac{P}{H} \) 

\(\cos\sec \theta = \frac{H}{P}\)
\(\cos \theta = \frac{B}{H}\)

\(\sec \theta = \frac{H}{B}\)

\(\tan \theta = \frac{P}{B}\) 

\(\cot \theta = \frac{B}{P}\)

Learning all these formulae is somewhat difficult because trigonometry is full of formulas. Let me tell you the very easy method to learn the above formulas.
Following is the basic pattern which you should learn by heart. The easy way to remember this is the way our teachers teach us in schools. I am sure that is the best way to remember this. You can never forget this trick:
(Some People Have) (Curly Brown Hair) (Through Proper Brushing)
Learn the above line and write 3-3 words vertically like:
\(\begin{array}{|c|c|}\hline \text{Some} & \text{Curly} & \text{Through} \\ \hline \text{People} & \text{Brown} & \text{Proper} \\ \hline \text{Have} & \text{Hair} & \text{Brushing} \\ \hline \end{array}\)
\(\begin{array}{|c|c|}\hline S & C & T \\ \hline P & B & P \\ \hline H & H & B \\ \hline \end{array}\)

\(\begin{array}{|c|c|}\hline \sin & \cos & \tan \\ \hline P & B & P \\ \hline H & H & B \\ \hline \cos\sec & \sec & \cot \\ \hline \end{array}\)

Now, It will seem very easy to learn the ratios. Isn't it?

Basic Formula

Following are some very basic formulae. Go through these formulas once, then we will proceed to the trigonometric ratios of standard angle.

\(\cos\sec \theta = \frac{1}{\sin \theta}\)

\(\sec \theta = \frac{1}{\cos \theta}\)

\(\cot \theta = 1\tan \theta\)

\(\tan \theta = (\sin \theta)\cos \theta\)

\(\sin^2\theta + \cos^2\theta = 1'\)

\(1+ \tan^2 \theta = \sec^2 \theta\)

\(1+ \cot^2 \theta = \cos\sec^2 \theta\)​​​​​​​

Trigonometric Ratios of Standard Angles

You must have seen the following table of trigonometric ratios, but so many times you may get confused by remembering the values. Here, I will tell you the basic technique which will help you to learn this table.

\(\begin{array}{|c|c|}\hline \theta & 0^{\circ} & 30^{\circ} & 45^{\circ} & 60^{\circ} & 90^{\circ} \\ \hline \sin \theta & 0 & \frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{\sqrt{3}}{2} & 1 \\ \hline \cos \theta & 1 & \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} & 0 \\ \hline \tan \theta & 0 & \frac{1}{\sqrt{3}} & 1 & \sqrt{3} & \text{Not Defined} \\ \hline \cot \theta & \text{Not Defined} & \sqrt{3} & 1 & \frac{1}{\sqrt{3}} & 0 \\ \hline \cos\sec \theta & 1 & \frac{2}{\sqrt{3}} & \sqrt{2} & 2 & \text{Not Defined} \\ \hline \sec \theta & \text{Not Defined} & 2 & \sqrt{2} & \frac{2}{\sqrt{2}} & 1 \\ \hline \end{array}\)

Write numbers from 0 to 4  Firstly, to find the values of Row1


Divide the numbers with 4

Take the square root of all

\(1 \to \frac{1}{4} \to \frac{1}{4}\)

\(2 \to \frac{2}{4} \to \frac{1}{2}\) Taking Square root\( \to \) \(\frac{1}{\sqrt{2}}\)

(\(3 \to \frac{3}{4} \to \frac{3}{4}    \)

\(4\to \frac{4}{4} \to 1  \)


See. how simple. You just need to divide 4 to numbers 0 to 4, then take square root.

Row2 = It is just the reverse of Sin values. (See table)

Row3 = Value of tan `theta` is derived by using (\((Sin \theta) cos\theta\))

Row4 = Use \((\cot \theta = \frac{1}{\tan\theta})\)

Row5 = Use \(\sec (\theta= \frac{1}{\cos \theta})\)

Row6 = Use \((\cos\sec \theta = \frac{1}{\sin \theta})\)

Examples with Solution

Example1: Find the value of:

\((\frac{2\tan30^{\circ}}{1+ \tan^2 30^{\circ}})\)

Solution:\((\frac{2\tan30^{\circ}}{1+ \tan^2 30^{\circ}} =\frac{2}{\sqrt3}1+\frac{1}{\sqrt3}^2)\)

Solving above will give = \(\sqrt3\)

Example2: If  \(\tan\theta= \frac{\sqrt2}{\sqrt3}\)`, then what will be the value of \(cos\theta\)?

Solution: \(\tan \theta = \frac{P}{B}\)

Therefore, \(P = \sqrt2\ and\ B = \sqrt3\)

using Pythagoras Theorem, \(H^2 = P^2 + B^2\)

\(H^2 = 2+ 3 = 5\)

⇒ \(H = \sqrt5\)

Therefore, \(\cos \theta = \frac{B}{H} = \frac{\sqrt3}{\sqrt5}\)

I hope you understand this topic till now, I will soon update the rest of the topic.
Your valuable feedback will be appreciated.

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