# 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.
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(𝜶 – θ)= ?
(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
𝜶 – θ = 600
As
we per the question, we need the value of-
tan
(𝜶 – θ), so substituting the value we get-
tan 600= ?
tan 600=√3
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

### 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