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

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