# Trigonometry III – Using Trigonometric Formulas to Solve Trigonometry Questions

### Trigonometric Formulas are the basics of solving trigonometry questions in the traditional way. Read on for a thorough understanding of trigonometry questions for SSC Exams.

Trigonometry Problems can be solved by the use of trigonometric formulas, however sometimes this conventional method can be time taking. In our series on Trigonometry we started with discussing- the basics of what is trigonometry, the important formulas and identities and have now come to the 3rd blog. In the 3rd blog of the series we’ll discuss some traditional and smart methods to solve trigonometry questions in SSC Exams.
Before we move ahead, it would be a good idea to quickly revise the important trigonometric formulas and identities.

### Set 1: Questions Based on Trigonometric Ratios and Identities

A set of Trigonometry Problems can be solved by using trigonometric ratios and identities.

Problem 1: sin2 250 + sin2 650 =?
a) √3/2      b) 1      c) 0      d) 2/√3

Solution 1:
We know the value of sin only for some specific angles like 300, 450, 600 and so on, so there is no point in trying to substitute the value. Instead we need to some trigonometric formulas or identities to solve such trigonometry questions.
sin2 250 + sin2 650 = ?
=> sin2 250 + sin2 (900 - 250)
We know that for the trigonometric ratio changes for-
=> sin2 250 + cos2 250
Using the above identity we get-
=> 1
Another way to solve this question is by converting sin2 250 to cos2 650, in this case also the approach will be similar.

Therefore the value of the above trigonometric expression is- Option b

Problem 2: cos4θ + sin4θ =2/5, then of 1- 2sin2θ =?

Solution 2:
Observing LHS of the equation below-
cos4θ + sin4θ =2/5
We find that it is in the form of-
Expressing LHS of the above trigonometric equation in this form we get-
(cos2θ + sin2θ) (cos2θ - sin2θ)= 2/5
Using the following trigonometric identity in the equation we get-
1 (cos2θ - sin2θ)= 2/5
Once again using the trigonometric identity => cos2θ + sin2θ = 1, in a different form-
Replacing that value we get-
=> 1- sin2θ - sin2θ = 2/5
=> 1- 2sin2θ - sin2θ = 2/5

The expression on the LHS is the same as the expression whose value we have to find.

### Set 2: Questions Based on Change in Trigonometric Ratios and Angles

A set of Trigonometry Problems can be solved by using trigonometric formulas where ratios change based on the value of angles.

Problem 1: If sin(600- θ) = cos(𝜶 - 300), then tan(𝜶θ)= ?
(𝜶  and θ are positive acute angles, where 𝜶> 600  and θ < 600
a) 1/√3     b) 0      c) √3      d) 1

Solution 1:
Taking LHS of the given equation and using one of the trigonometric formulas for the same we get-
=> sin(600- θ)= sin[900- (𝜶 - 300)]
Since the trigonometric ratio on both the ends is the same, we can equate the angles-
600 - θ= 900- 𝜶 + 300
𝜶θ = 600

As we per the question, we need the value of-
tan (𝜶θ), so substituting the value we get-
tan 600= ?
tan 600=√3

Therefore the answer is- Option c

Therefore such questions can be solved by using the appropriate trigonometric formulas.

### Set 3: Questions Based on converting one set of Trigonometric Ratios in another set of Trigonometric Ratios

A set of Trigonometry Problems can be solved by using trigonometric formulas and identities.

Problem 1: (tan 570 + cot 370)/(tan 330 + cot 530) = ?
a) tan 330cot 530    b) tan 530cot 330      c) tan 330cot 570      d) tan 570cot 370

Solution 1:
Let’s try and convert these trigonometric ratios into something else and use trigonometric formulas to finally reduce them to one of the answer options.

We know-
=> tan 570 = cot 330  (i)
=> cot 370 = tan 530  (ii)

And we convert ratios in the denominator to their reciprocal ratios, doing this and using (i) and (ii) we get-
=> cot 330 +   tan 530 / [(1/cot 330) + (1/tan530)

The idea behind doing this was to get similar ratios and angle values in both the numerator and the denominator. Simplifying this we get-
=> cot 330 +   tan 530 / [(cot 330 +   tan 530) cot 330 tan530]
=> cot 330 tan530

Therefore the answer is- Option b

### Practice Questions Based on Trigonometric Formulas

Question 1: If x = cosec 𝜃 – sin 𝜃 and y = sec 𝜃 – cos𝜃 then the value of x2y2 (x2y2 + 3) is?
a) 0       b) 1       c) 2       d) 3

Question 2: If sin 𝜃 + sin2𝜃 = 1, then the value of = cos12𝜃 + 3 cos10𝜃 + 3cos8𝜃 + cos6𝜃 – 1 is?
a) 0       b) 1       c) -1       d) 2

Don’t forget to write your answers in the comment section and do tell us how this series on trigonometry helped you understand the topic!

Remember to keep practicing!