###
**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 3

^{rd}blog. In the 3^{rd}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:**sin

^{2}25

^{0}+ sin

^{2}65

^{0}=?

a) √3/2 b) 1 c) 0
d) 2/√3

**Solution 1:**

We
know the value of sin only for some specific angles like 30

^{0}, 45^{0}, 60^{0 }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.
sin

^{2}25^{0}+ sin^{2}65^{0}= ?
=>
sin

^{2}25^{0}+ sin^{2}(90^{0}- 25^{0})
We
know that for the trigonometric ratio changes for-

=>
sin

^{2}25^{0}+ cos^{2}25^{0}
Using
the above identity we get-

=>
1

Another
way to solve this question is by converting sin

^{2}25^{0}to cos^{2}65^{0}, in this case also the approach will be similar.
Therefore
the value of the above trigonometric expression is- Option b

**Problem 2:**cos

^{4}θ + sin

^{4}θ =2/5, then of 1- 2sin

^{2}θ =?

**Solution 2:**

Observing
LHS of the equation below-

cos

^{4}θ + sin^{4}θ =2/5
We
find that it is in the form of-

Expressing
LHS of the above trigonometric equation in this form we get-

(cos

^{2}θ + sin^{2}θ) (cos^{2}θ - sin^{2}θ)= 2/5
Using the following trigonometric
identity in the equation we get-

1
(cos

^{2}θ - sin^{2}θ)= 2/5
Once again using the trigonometric
identity => cos

^{2}θ + sin^{2}θ = 1, in a different form-
Replacing that value we get-

=>
1- sin

^{2}θ - sin^{2}θ = 2/5
=>
1- 2sin

^{2}θ - sin^{2}θ = 2/5
The expression on the LHS is the same
as the expression whose value we have to find.

Therefore the answer is- 2/5

###
**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(60

^{0}- θ) = cos(𝜶 - 30

^{0}), then tan(𝜶 – θ)= ?

(𝜶 and θ
are positive acute angles, where 𝜶> 60

^{0}and θ < 60^{0}
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(60

^{0}- θ)= sin[90^{0}- (𝜶 - 30^{0})]
Since
the trigonometric ratio on both the ends is the same, we can equate the angles-

60

^{0 }- θ= 90^{0}- 𝜶 + 30^{0}
𝜶 – θ = 60

^{0}
As
we per the question, we need the value of-

tan
(𝜶 – θ), so substituting the value we get-

tan 60

^{0}= ?
tan 60

^{0}=√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 57

^{0}+ cot 37

^{0})/(tan 33

^{0}+ cot 53

^{0}) = ?

a) tan 33

^{0}cot 53^{0}b) tan 53^{0}cot 33^{0}c) tan 33^{0}cot 57^{0}d) tan 57^{0}cot 37^{0}

**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 57

^{0}= cot 33^{0 }(i)
=>
cot 37

^{0}= tan 53^{0 }(ii)
And
we convert ratios in the denominator to their reciprocal ratios, doing this and
using (i) and (ii) we get-

=>
cot 33

^{0 }+^{ }tan 53^{0}/ [(1/cot 33^{0}) + (1/tan53^{0})
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 33

^{0 }+^{ }tan 53^{0}/ [(cot 33^{0 }+^{ }tan 53^{0}) cot 33^{0}tan53^{0}]
=>
cot 33

^{0}tan53^{0}
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 x

^{2}y^{2}(x^{2}y^{2 }+ 3) is?
a) 0 b) 1 c) 2 d) 3

Question 2: If sin 𝜃 + sin2𝜃
= 1, then the value of = cos12𝜃
+ 3 cos

^{10}𝜃 + 3cos^{8}𝜃 + cos^{6}𝜃 – 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!

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