# Easy Method to Solve Maximum and Minimum Values in Trignometry For SSC Exams

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**Basic Methods - Effective For SSC CGL, SSC CHSL, SSC MTS, UPSC Prelims.**

*Today we will discuss about some basic short method, How to solve Minimum and Maximum values in Trignometry for SSC CGL Exams. As we have seen 2-3 questions was generally asked from this section. Many people finds it difficult to solve this types of problem. Today Just note-down the basic formula's. It will help you to understand the logic behind it. Just Apply the tricks.*

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**Basic Formula's**
As you all know the basic formula’s in Trigonometric Identities -

- $Si{{n}^{2}}\theta +Co{{s}^{2}}\theta =1$
- $1+Ta{{n}^{2}}\theta =Se{{c}^{2}}\theta $
- $1+Co{{t}^{2}}\theta =Co{{\sec }^{2}}\theta $

**Basic Formula Chart**

**Basic Rules and Formula’s to Understand the Logic Behind it & Apply it on your questions-**

Another Rules to
Solve By using A.P & G.P –

Arithmetic Mean =
$\frac{a+b}{2}$

Geometric Mean =
$\sqrt{ab}$

Now Lets see the
examples – Here we will discuss those questions only which was asked in
previous papers of SSC CGL, CHSL Only.

**Example 1. Find the Minimum Values of $Se{{c}^{2}}\theta +Co{{\sec }^{2}}\theta =?$**

**Solution- If You find any question like this i.e. Second term is the reciprocal of the first then try to convert your questions according to the chart.**

**Here we will going to convert & apply some basic Formula’s from the above –**

**As we know**

$1+{{\tan
}^{2}}\theta =Se{{c}^{2}}\theta $

$1+Co{{t}^{2}}\theta
=Co{{\sec }^{2}}\theta $

Applying Both the
Basic Formula’s, We Get –

2+${{\tan
}^{2}}\theta +Co{{t}^{2}}\theta $.

Now Apply for the
above Chart- We get the formula for minimum values, Apply
on it

2+${{\tan
}^{2}}\theta +Co{{t}^{2}}\theta$= 2+ $2\sqrt{1*1}$

= 4 Answer !!

**Examples 2. Find the Minimum values of $Si{{n}^{2}}\theta +Co{{s}^{2}}\theta +Ta{{n}^{2}}\theta +Co{{t}^{2}}\theta +Co{{\sec }^{2}}\theta +Se{{c}^{2}}\theta =?$**

**Solution – As you can see in the questions & we know that**

$Si{{n}^{2}}\theta +Co{{s}^{2}}\theta =1$

Hence applying the above formula we get,

$1+{{\sec
}^{2}}\theta +\cos e{{c}^{2}}\theta +{{\tan }^{2}}\theta +{{\cot }^{2}}\theta $

Now we know (Using A.P & G.P Formula)-

${{\tan }^{2}}\theta +{{\cot }^{2}}\theta =2$

Therefore,

$1+2+{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta
$

Now Change ${{\sec }^{2}}\theta ,\cos
e{{c}^{2}}\theta $

$1+2+\frac{{{\sin }^{2}}\theta +{{\cos
}^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }$

Now again Apply – the basic formula from the
chart

Hence finally we get –

$1+2+\frac{1}{{{(\frac{1}{2})}^{2}}}$

=7 Answer !!

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