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In this section, we’ll study the basics of trigonometry and its identities.

TRIGONOMETRY Trigonometry is basically the study of length and angle of a triangle. There are 6 trigonometric identities we come across while learning about its concepts. It includes:

1. Sine

2. Cosine

3. Tangent

4. Cosecant

5. Secant

6. Cotangent

All of these are abbreviated as sin, cos, tan, cosec, sec, and cot.

All these are called Trigonometric Identities.

They all have a relationship with the sides of the triangle. Let’s look at them –

To study the trigonometric function we use a right-angled triangle. Let’s look below:-

Let ABC be a triangle right angled at A and angle B=θ, P is called perpendicular, b is for base and h is the hypotenuse.

So, we’ll look the trigonometric functions :

Here, sinθ = P/h

cosecθ= h/P

cosθ= b/h

secθ= h/b

tanθ = p/b

cotθ= b/P

What we observe here is that cosecθ = 1/sinθ , secθ=1/cosθ,cotθ=1/tanθ.

This is interesting to note in trigonometry!!!

In the beginning, you may have some difficulties in grasping these. So, there is a quick trick to remember it.

So you’ll go like “Pain before Practice, Hair hair Black”. The very first letter of each word represents either side (P, b, h) and from the table, you can easily find which trigonometric function corresponds to which.

As these, all are related to a Pythagorean triangle that’s why they are called Pythagorean identity.

Was that difficult? I don’t think so and hope you too.

Let’s look at the example below to understand it better and how can we solve these questions as per the exam perspective.

e.g., If tan θ + cot θ = 16, then find the ratio of tan2θ + cot2θ + 20 tan θ. cot θ?

Sol:-  As we have been given here tan θ+ cot θ= 16, we need

tan2θ + cot2θ. So, how we are going to get it? Obviously by squaring on both sides. So we get –

(tan θ+ cot θ)2= 162

tan2 θ +cot2 θ + 2 tan θ. cot θ = 256

and as we know tan θ=1/ cot θ

we will see the term 2 tan θ. cot θ reduces to 2

so, we have tan2 θ+ cot2 θ = 256-2 =254

and we wish to evaluate

tan2θ + cot2θ + 20 tan θ. cot θ = 254 + 20 tan θ. cot θ

= 254+ 20              ( as tan θ=1/cot θ)

=274

It was easy, isn’t? Practice more and more.

Let’s now move forward with the values of the functions in the quadrant.

What I actually mean by the values is the sign of the functions in different quadrants. As we move from quadrant-to-quadrant signs changes for different trigonometric functions.

VALUES OF FUNCTIONS IN A QUADRANT

You can easily see the signs of different functions on the different quadrants.

This is an important rule of trigonometry.

If you can’t remember it there is a short trick!!

All you need to learn this A means all are positive, S means only sin and its reciprocal cosec is positive, T means tan and cot are positive and C means cos and sec are positive. Simple? But remember it’s a secret!!!

Ques:- If sin θ =8/17  and 90° < θ < 180°, then find the value of the expression 2sin θ+cos θ/3cos θ+5sin θ? (SSC-CGL 2017)

Solution: since θ lies in the second quadrant and we know cos θ is negative here.

Using Pythagoras identity,

Here we see sin θ =8/17 and from our learning, above we know.

sin θ =P/h =8/17. let us take a triangle ABC whose Perpendicular is 8 and hypotenuse is 17, so using Pythagoras theorem base= 15 ( as 172 – 82= 152)

As you’ll come across the values of these trigonometric functions at different angles. It becomes important for us to know what these values are. The table below gives you the clear data.

Note:- The signs of the values of each function changes as we move to different quadrants so the values also change.

There are some formulas that we come across while we solve our question for SSC-CGL QUANT. Let us look at them to make our life simpler { which really never gets simple(crying in the corner)}

FORMULAE

There is actually a long list of formulas so be prepared. I shall now begin.

To make our Calculations simpler we have some relations between these trigonometric functions.

1. sin2θ + cos2θ= 1

2. 1+ tan2θ= sec2θ

3. 1+ cot2θ= cosec2θ

These are the general formulas that help you to easily access to the solutions. Once you have grasped them in your mind feels like you have accomplished something { hurray!!}

As it always said quant requires practice. So we are going to practice previous years’ questions asked in SSC-CGL EXAM.

Ques:- If cos θ = -1/2   and π  < θ < 3 π  , find the value of 4tan2 θ – 3 cosec2 θ?

Solution:- We have been given cos θ=-1/2

As we need to find the value of tan θ which is equal to   sin θ /cosθ and cosec θ  =1/sinθ

From the formula, we just learned sin2θ + cos2θ= 1

So, sinθ= ±√1-cos2θ

We have the value of cosθ=-1/2

sinθ= ±√1-(-1/2)2 = ±√1-(1/4) = ±√3/4 =±√3 / 2

now, we will check what is the sign of sinθ, as θ lies π  < θ < 3 π

it means it is in the third quadrant and we know from the above chart that sinθ is negative here.

So we have   sinθ=- √3 / 2

We will now calculate the value of tanθ and cosec θ

Tan θ= sin θ /cosθ = – √3 / 2 / -1/2 = -√3

Cosecθ = 1/ sinθ = -2 / √3

But our ques asked us to find the value of 4tan2 θ – 3 cosec2 θ

We will just simply substitute the value of tanθ and cosecθ

Therefore,

4tan2 θ – 3 cosec2 θ= 4 ×(-√3)2– 3 (-2 / √3) 2

= 4×3 – 4

= 12-4 = 8

Wasn’t it easy? Just have your concepts clear.

Ques:- If 2 sin 2θ – √3  = 0, then the value of Θ lies between?

Sol: Since, we have, 2 sin2θ – √3  = 0

sin2θ = √3/ 2

Look at the table of trigonometric values, where do we have the value of sin √3/ 2?  At 60 o.

So this becomes equals to sin 60 o.

sin2θ= sin 60o

=>          2θ = 60o

=>             θ= 30o

As 30 lies in first quadrant.  So 0< θ< π/2.

Let’s see the next question .

Ques:- If 8 sin x = 4 + cos x, the find the values of sin x?

Sol:- as we know, sin2θ + cos2θ= 1

cosθ = √1- sin2θ

Now,   8 sin x – 4 = cos x

⇒ 8 sin x – 4 = √1- sin2x

Squaring on both sides,

⇒ 1 – sin2x = (8 sin x – 4)2

⇒ 1 – sin2x = 64 sin2x + 16 – 64 sin x

⇒ 65 sin2x -64 sin x + 15 = 0

As this is a quadratic eq. we do the middle term splitting

⇒ 65 sin2 x – 39 sin x  – 25 sin x + 15 = 0

⇒ 13 sin x (5 sin x  -3)  –  5 (5 sin x  –  3) = 0

⇒ (5 sin x – 3) (13 sin x – 5) =0

⇒ sin x =  3/5  or sin x =5/13.

SO, this is what actual ques are.