HOW TO SOLVE HARD AND IMPOSSIBLE PUZZLES PART 2

 we are going to discuss some of the puzzles, challenges or brain teaser which are viral in social media such as Face book ,whats app and twitter    etc. which are based on Mathematics directly or indirectly . You have seen such types of puzzles mentioned 99.9% Fail , Solve if you are genius like this........

Let us try to discuss and solve them. 

PUZZLE # 1

Which number should replace the ?

  4 ,  9 ,  19  ,  39  , 79   ?

when we subtract   4 from  9  we get   5 i.e     9 -  4  =  5 

when we subtract   9 from 19 we get 10 i.e   19 -  9  = 10 

when we subtract 19 from 39 we get 20 i.e.  39 - 19 = 20

when we subtract 39 from 79 we get 40 i.e.  79 - 39 = 40

Now analyse the trends of answers ,we are getting double of the previous number as a result every time .Similarly if we follow the same trend from 79 to next term ,then we shall get 80 as answer. So to get 80 as answer "Add 80 to 79 to get 159 ". when we subtract 79 from 159 ,we get the double of the previous result .

So final answer is  159 - 79 = 80.

  4 ,  9 ,  19  ,  39  , 79  , 159

 Therefore  " ? " will replace 159 as Answer.

PUZZLE # 2

But before we proceed to discuss the Puzzle-1, let us discuss some of the necessary keywords/definitions which have been used in  it.

Even Number

When any Number is divisible by  2 then that number is called An Even number. If we take 2, 16, 84, 22, 20, 100 etc then they are Even Numbers as they can be divided by 2.

Let us learn Multiplication , division , arithmetic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics.

Odd Number

When any number is NOT divisible by  2 then that number is called An ODD number .If we take 21, 15, 49, 21, 57, 101 etc then they are Odd  Numbers as they can NOT  be divided by 2.

Multiple of 3 

An number which is divisible by 3 or which comes in the  table of 3 is called  multiple of 3. if we take 57, 102, 69, 9, 6, 21, 39 etc numbers , as these numbers comes in the table of 3 or these are divisible by 3 , so they are Multiple of 3.

Prime Number

Any number which has only two Factors 1 and the numbers itself is called Prime number. 1 is not the the prime number, 2 ( It is the 1st prime number ) , 3, 5, 7, 11, 13, 17, 19, 23  etc  are examples of Prime numbers as these numbers have only two factors 1 and the number itself.

Square Number

Any number whose square root is a Natural  number is called square Number . such as  1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 etc.

 Factors of 360

All those numbers which can divide 360 are called factors of 360 .Here We can take such numbers as 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 18, 36 etc

How to Solve 

Have you still not visited "HOW TO SOLVE HARD AND IMPOSSIBLE PUZZLES  PART 1"

PUZZLE # 3

When we multiply   2   with  4,  we get     8
When we multiply   3   with  6,  we get    18
When we multiply   4   with  8,  we get    32
When we multiply   5   with 10, we get    50
When we multiply   6   with 12, we get    72
When we multiply   7   with 14, we get    98
When we multiply   8   with 16, we get    128
When we multiply   9   with 18, we get    162
When we multiply  10  with 20, we get    200

Analyse the Answers above  , we get  4, 6, 8, 10, 12, 14, 16, 18 and 20. Is there any relation between them. Of course, Because every answer is 2 more than the previous Answer. So Last number  should be 200. so in order to get answer 200 we have to multiply  20 with 10.

So       If 2  =  8  

               3  = 18 

               4  = 32

               5  = 50 
               6  = 72  
               7  = 84
               8  = 128
               9  = 172 
              10 = 200 

Then    10 = 200

In order to get any number any of any term  based upon this trends we can use the formula  2n² , where "n "  is the number appearing on left side of the Puzzle.

So Final Answer will be 200

Here is the Solution of the Puzzle asked in "HOW TO SOLVE HARD AND IMPOSSIBLE PUZZLES PART 3" as follows.

There are two numbers 'a' and 'b' in 1st and 2nd columns then 3rd column has been  occupied by their combination as follows

a*b = (a²  + b² ) - 1,

1*3 = (1²  + 3² ) - 1 = ( 1 +   9) -1  = 10 - 1 =  9

3*4 = (3²  + 4² ) - 1 = ( 9 + 16) -1  = 25 - 1 = 24

2*5 = (2²  + 5² ) - 1 = ( 4 + 25) -1  = 29 - 1 = 28

4*5 = (4²  + 5² ) - 1 = (16 + 25) -1 = 41 - 1 = 40

So 40 will replace  ?  in this puzzle   

Conclusion

Thanks for devoting your valuable time for this post  "How   to solve various Hard and impossible puzzles ,Quizzes ,  Brain Teasers and challenges " of my blog . If you liked this this blog/post,  Do Follow me on my blog and share this post with your friends . We shall meet again   in next post with solutions of most interesting and mind blowing puzzles ,till then Good Bye.

                                                                       

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HOW TO SOLVE IMPOSSIBLE AND HARD PUZZLES ,QUIZZES PART 1

HOW TO SOLVE IMPOSSIBLE AND HARD PUZZLES

Hello Friends Most welcome to this post of my Blog . In this post We are going to discuss some of the puzzles, quizzes , riddles, challenges and brain teaser which are viral in social media such as Face book ,whats app, twitter and Google + etc. which are based on Mathematics directly or indirectly . You also have seen such types of problems mentioned 99.9% Fail , Solve if you are genius like this...

Let us try to discuss and solve some of these Puzzles one by one.

Puzzle #1 

If 
2 = 6
3 = 12
4 = 20
5 = 30
6 = 42
then 9 = ?
In this problem two steps are missing 
7= ?
8 = ?
and directly asked how 9 = ?

HOW TO SOLVE IMPOSSIBLE AND HARD PUZZLES

How 2 is related to 6 , in the same manners 3 is related to 12 and 4 is also related to 20  and so on.


If we  multiply   2 with its successor ( the immediate  next number) i.e. 3 then we get 2×3 = 6,  Similarly  multiply 3 with its successor 4 we get 3 ×4 =12 ,4×5 =20  and 5 ×6 =30 ,6×7=42 and ( and the missed numbers 7×8=56,8×9=72 ) 


and finally we have a formula for this quiz which is (a)×(a+1) ,where  "a" is given number.

The required answer    is 9 ×10 = 90

Read  " How to Multiply Two numbers in quick Time " to  save your time in calculations

Now let us move to little bit harder problem below

Puzzle #2

HOW TO SOLVE IMPOSSIBLE AND HARD PUZZLES

Look at this problem , To solve such types of puzzles ,1st of all do not count "+" ,"-" or any mathematical sign as it appears. i.e in such puzzles + does not means addition and also  - do not means subtraction. we our self have to make a formula to satisfies given results and conditions we have to perform such mathematical  operations whose result must be the number given in the right side .

So if we add 20+20 which is 40 and then divide it with 10 we shall get 4,

Similarly So if we add 15+15 which is 30 and then divide it with 10 we shall get 3,

so at last if we add 30+15 which is 45 and then divide it with 10 we shall get   4.5  as answer.

So the formula for this quiz is a+b =  (a+b)/10
---------------------------------------------------- ------------------------------------------------ -----------------------

Puzzle # 3 

Here it is given that 6+9 =61
5+8 = 46
4+7 = 33
3+6 = 22
2+1 = Missing
1+4 = ?

Most of peoples forget to understand that one step has been left intentionally to  make us fool or we can say it is the part of their strategy.

HOW TO SOLVE IMPOSSIBLE AND HARD PUZZLES

Step 1

If we  Multiply   6 ×9 then it is equal to 54 but not equal to the given result 61, 

Step 2

If we add successor of 1st term to the result just obtained then it  is equal to 54+(6+1) = 61. yes it is , but is it true for  next step.

5×8=40 , adding 1st term's successor to the result just obtained i.e 6

5×8+(5+1)=40+6=46, yes it works

Now let us verify for remaining steps
4×7+(4+1) = 28+5 =33 , ok
3×6+(3+1)=18+4=22 ,  it is ok for all given results
So we have discovered the trends 
1 Multiply    both the given numbers
2 Add the successor of the 1st number to the result obtained in step 1 ,and we have our answers every time in this puzzles, 
.
.
So using the formula so discovered the required term must be  1×4+(1+1)=4+2 =6

Now the formula for this puzzle is a+b = (a×b)+(a+1)


Let us learn Multiplication , division , arithmatic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .

 Conclusion

Thanks for devoting your valuable time for this post  "How   to solve various puzzles ,Quizzes ,  Brain Teasers and challenges " of my blog . If you liked this this blog/post,  Do Follow me on my blog and share this post with your friends . We shall meet again   in next post with solutions of most interesting and mind blowing puzzles ,till then Good Bye.

Let us learn Multiplication , division , arithmetic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .

Magical   Maths

                                                                          

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Memorise A B AND C D Formulas In Trigonometry in an Easy Manner

Welcome to this post of learning trigonometric formulas  .Most of the Students or Mathematics Learner ,most of the time confuse to remember or memorise A  B and C  D formulas, They  mixed A B and C D formulas with each other and could not reproduce what they have learnt . So today we going to learn new techniques to learn "How to memorise AB and CD formulas" forever. Before this we must have knowledge of different trigonometric values of different angles in different quadrants.

First of all have a quick look at some of  these formulas

AB Formulas

S N	Fomula	Result
1	sin (A + B)	sin A cos B + cos A sin B
2	sin (A - B)	sin A cos B - cos A sin B
3	cos (A + B)	cos A cos B - sin A sin B
4	cos (A - B)	cos A cos B + sin A sin B

To clear all your doubts on   " How to Calculate Different Trigonometric values in different quadrants "  in an easy Method. Click on the  above  links  .

Tricks to Learn    A  B   Formulae  For  sine  angles

When angles are added  i. e  Sin  ( A+B )  

When Angles are added and then their Trigonometric Ratios is taken , and if we have to take the  Sine of  added angles, then it can be done like this.

Start with  sine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign      Start with Cosine of angle A and multiply with Sine of angle B. i.e. start with sine and ends with sine and in middle both the terms are cosine ,and angles start  A then B again A then again B.

Sin (A+B) = Sin A Cos B + Cos A Sin B

When angles are subtracted    i. e  Sin  ( A-B )  

When Angles are subtracted and  their Trigonometric Ratios is taken , and if we have to take the  Sine of  subtracted  angles, then it can be done like this

    
Start with  sine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign    Start with Cosine of angle A and multiply with Sine of angle B. i.e  start with sine and ends with sine and in middle both the terms are cosine ,and angles start  with A then B again A then again B.

Sin (A - B) = Sin A Cos B - Cos A Sin B

Tricks to Learn    A  B   Formulae For  Cosine  angles

When angles are added   i. e  Cos  ( A+B )  

When Angles are added and then their Trigonometric Ratios is taken , and if we have to take the  Cosine  of  added angles, then it can be done like this.

Start with  cosine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  -ve sign      Start with Sine of angle A and multiply with Sine of angle B. i.e. 1^st   and 2^nd terms are   cosine and  3^rd and 4^th terms    are sine , Angles start with   A then B again A then again B.

"Here  Sum of cosine of  Two angles  is equal to difference of  product of  cosines of both the angles    and product of sine of both the angles ".

Cos (A+B) = Cos A Cos B - Sin A Sin B

When angles are subtracted    i. e  Cos  ( A-B )  

When Angles are subtracted and  then their Trigonometric Ratios is taken , and if we have to take the  cosine of  subtracted  angles, then it can be done like this.

Cos (A - B) = Cos A Cos B + Sin A Sin B

Start with  cosine of  angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign      Start with Sine of angle A and multiply with Sine of angle B. i. 1^st   and 2^nd terms are   cosine and  3^rd and 4^th  terms    are sine , and angles start with   A then B again A then again B.

"Here  Difference  of cosine of  Two angles  is equal to the  Sum  of  product of  cosines of both the angles    and product of sine of both the angles" .

Want to Learn  What is Set , Types of Sets , Union,  Intersection of Sets  And Venn Diagrams  

How to Memorise C D Formulae

To learn C D formulae 

Step 1 

Place 2 for all four formulae and  take Trigonometric Ratio of 1st angle for all four formulae which  is (C+D)/2 and again  trigonometric Ratio of  2nd angle which  is (C-D)/2.

Step 2.1 

For addition of Sine Formula start with sine of 1st angle as mentioned in step 1 and multiply it with cos of  2nd angle as mentioned in step 1.

Step 2.2 

For subtraction of Sine Formula start with cosine of 1st angle as mentioned in step 1 and multiply it with sine of  2nd angle as mentioned in step 1.

Step 3.1

For addition of cosine Formula start with cosine of 1st angle as mentioned in step 1 and multiply it with cosine  of  2nd angle as mentioned in step 1.

 

Step 3.2

For subtraction of cosine Formula start with sine of 1st angle as mentioned in step 1 and multiply it with sine   of  2nd angle as mentioned in step 1,and do not forget to multiply it with -ve sign.

or  

If you do not want to multiply it with -ve sign  ,then you can change 2nd angle (D-C)/2 instead of (C-D)/2

CD Formulas

S N	Fomula	Result
1	sin C + sin D	2 sin {(C+D)/2} cos {(C-D)/2}
2	sin C - sin D	2 cos {(C+D)/2} sin {(C-D)/2}
3	cos C + cos D	2 cos {(C+D)/2} cos {(C-D)/2}
4	cos C - cos D	-2 sin {(C+D)/2}  sin {(C-D)/2}
or	cos C - cos D	2 sin {(C+D)/2}   sin {(D-C)/2}

Want to learn Matrix , Different Types of matrices And Determinants.  

Want to learn How to Solve Linear Equations of Two and Three Variables  || Matrix Method to Solve System of Linear Equations.

A quiz of mathematics for you

Conclusion

Thanks for devoting your valuable time for the post Easy Tricks to Memorise A B and C D Formulae in Trigonometry and Trigonometry 's shortcut formulas of this blog ,trigonometry formulas for class 11 ncert,trigonometric functions class 11 notes, trigonometry class 11 tricks,,trigonometry formulas list,. If you found this this blog/post of your concern, Do Follow me on my blog and share this post with your friends . We shall meet again in next post ,till then Good Bye.

Let us learn Multiplication , division , Logarithmaetic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .

How to Memorise A   B and C   D  formulas  easily ,watch this video 

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HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES

Welcome to this post of learning trigonometric formulas in an easy way with me. Because most of the Students or Mathematics Learner ,most of the time confuse to remember or memorise value of different trigonometric angles in different quadrants and could not reproduce what they have learnt . So today we are  going to learn new techniques to learn "How to memorise different values of trigonometric angles in various quadrants" forever. Before this we must have knowledge of different trigonometric values of different angles in different quadrants.

When angle lies in 1st Quadrant

(1) When angle lies in 1st quadrant, then all the t- Ratios have positive values. As in 1st quadrant all the three  sides Perpendicular ,base and hypotenuse of  right angled triangle are positive.

      

(2) When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan  x to cot x , cos x to sin x ,cosec c to sec x and sec  x to cosec x.

MEMORISING TRIGONOMETRIC ANGLES

 Table for  Trigonometric Angles  in 1st Quadrant 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (π/2 - x)	cos x
2	Cos (π/2 - x)	sin x
3	tan (π/2 - x)	cot x
4	Cot (π/2 - x)	tan x
5	Sec (π/2 - x)	Cosec x
6	Cosec (π/2 - x)	Sec x

(3)  When angle involve  2π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x  to sin x  , cos x  to cos x  , tan x  to tan  x and so on.

 Table for  Trigonometric Angles  in 1st Quadrant 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (2π + x)	Sin x
2	Cos (2π + x)	cos x
3	tan (2π + x)	tan x
4	Cot (2π + x)	Cot x
5	Sec (2π + x)	Sec x
6	Cosec (2π + x)	Cosec x

 

ALSO READ   MATRIX ,  DIFFERENT TYPES OF MATRICES AND DETERMINANTS .

When angle lies in 2nd  Quadrant

(1)  when angle lies in 2nd quadrants ,then only two t - ratios sin x and it reciprocal cosec x shall have +ve values and remaining t-Ratios shall have -ve values. As in 2nd  quadrant two  out of the three  sides Perpendicular and  hypotenuse of  right angled triangle are positive  while base is negative. So in all those  t -ratios ,when base involves  they will be negative. So  cos x, tan x ,cot x ,sec x involve with -ve value of base therefore these t- ratios shall be negative.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x to cosec x.

MEMORISING TRIGONOMETRIC ANGLES

 

Table for  Trigonometric Angles  in 2nd Quadrant 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (π/2 + x)	cos x
2	Cos (π/2 + x)	-sin x
3	tan (π/2 + x)	-cot x
4	Cot (π/2 + x)	-tan x
5	Sec (π/2 + x)	cosec x
6	Cosec (π/2 + x)	sec x

(3) When angle  involve  π ,then T - Ratios on the left hand side will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

 Table for  Trigonometric Angles  in 2nd Quadrant 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (π - x)	Sin x
2	Cos (π - x)	- cos x
3	tan (π - x)	-tan x
4	Cot (π - x)	-Cot x
5	Sec (π - x)	-Sec x
6	Cosec (π - x)	Cosec x

  

ALSO READ EASY TRICKS TO MEMORISE A B AND C D FORMULAS IN TRIGONOMETRY .

When angle lies in 3rd Quadrant

(1)  when angle lies in 3rd quadrants ,then only two t - ratios tan x and it reciprocal cot x shall have +ve values and remaining t-Ratios shall have -ve values. As in 3rd  quadrant two  out of the three  sides perpendicular and  base of  right angled triangle are negative  and hypotenuse is positive. So in all those  t -ratios ,when one of perpendicular or base  involves  they will be negative. So  sin x, cos x , sec x ,cosec x involve with -ve value of base or perpendicular therefore these t- ratios shall be negative. And tan x and cot x involves with both -ve values of perpendicular and base so they are positive in 3rd quadrant.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x  to cosec x.

MEMORISING TRIGONOMETRIC ANGLES

Table for  Trigonometric Angles  in 3rd Quadrant 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (3π/2 - x)	-Cos x
2	Cos (3π/2 - x)	-Sin x
3	tan (3π/2 - x)	cot x
4	Cot (3π/2 - x)	tan x
5	Sec (3π/2 - x)	-Cosec x
6	Cosec (3π/2 - x)	-Sec x

 
(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

Table for  Trigonometric Angles  in 3rd Quadrant 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (π + x)	-Sin x
2	Cos (π + x)	- cos x
3	tan (π + x)	tan x
4	Cot (π + x)	Cot x
5	Sec (π + x)	-Sec x
6	Cosec (π + x)	-Cosec x

 

 Let us understand these learning of trigonometric formulae with the help of this video

 

When angle lies in 4th Quadrant

(1)  when angle lies in 4th quadrants ,then only two t - ratios cos x and it reciprocal sec x shall have +ve values and remaining t-Ratios shall have -ve values. As in 4th quadrant two out of the three sides Base and hypotenuse of right angled triangle are positive and perpendicular is negative. So in all those t -ratios ,when perpendicular involves they will be negative. So sin x, tan x ,cot x ,cosec x involve with -ve value of perpendicular therefore these t- ratios shall be negative.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x  to cosec x.

MEMORISING TRIGONOMETRIC ANGLES

Table for  Trigonometric Angles  in 4th Quadrant 

 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (3π/2 + x)	-Cos x
2	Cos (3π/2 + x)	Sin x
3	tan (3π/2 + x)	-cot x
4	Cot (3π/2 + x)	-tan x
5	Sec (3π/2 + x)	Cosec x
6	Cosec (3π/2 + x)	-Sec x

So if we want to calculate sin 300° ,sin 240° and sin 330°  then it can be find out as follows

sin 300° = sin (270° + 30° ) = - cos 30° = -√3/2

sin 330° = sin (360° - 30° )  = - sin 30° = -1/2

sin 240° = sin (270° - 30° ) = - cos 30° = -√3/2

and if we want to calculate cos 300° , cos 240° and   cos 330°  then it can be find out as follows

cos 300° = cos (270° + 30° ) =  sin 30° = 1/2

cos 330° = cos (360° - 30° )  =  cos 30° = √3/2

cos 240° = cos (270° - 30° ) = - sin 30° = 1/2

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

Table for  Trigonometric Angles  in 4th Quadrant 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (2π - x)	-Sin x
2	Cos (2π - x)	cos x
3	tan (2π - x)	-tan x
4	Cot (2π - x)	-Cot x
5	Sec (2π - x)	Sec x
6	Cosec (2π - x)	-Cosec x

When Angle do not involve π 

Table for  Trigonometric Angles  in 4th Quadrant 

S N	Trigonometric    Angles	Value of T  Angles
1	Sin (- x)	-Sin x
2	Cos (- x)	cos x
3	tan (- x)	-tan x
4	Cot (- x)	-Cot x
5	Sec (- x)	Sec x
6	Cosec (- x)	-Cosec x

ALSO READ HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES || MATRIX METHOD TO SOLVE SYSTEM LINEAR EQUATIONS .

GENERALISATION OF THE FORMULAE

So when an angle involves integral multiple of π , i,e -3π, -2π, -π, 2π, 3π, 4π then of the T-Ratios will change , But + or - sign can be added at beginning , e. g . sin(nπ + x) may change to + sin x or - sin x ,similarly cos (nπ + x) may change to + or - cos x depending upon the quadrant in which angle lies.


So if we want to find the value of sin 1110° ,then it can be written as sin (3×360° + 30°) = sin 30° = 1/2 . (As the angle is lying in 1st Quadrant )

Similarly if we want to find the value of sin 1050° ,then it can be written as sin (3×360° - 30°) = - sin 30° = - 1/2. (As the angle is lying in 4th Quadrant )


and when an angle involves odd integral multiple of π/2 i.e. (2n+1)π/2 , i . e -7π/2 , -5π/2 , -3π/2 , π/2 , 3π/2 , 5π/2.

Conclusion

Thanks for devoting your valuable time for the post memorising different values of trigonometric angles in different quadrants of this blog. If you found this  blog/post of your concern, Do Follow me on my blog and share  post with your friends . We shall meet again in next post ,till then Good Bye.

 

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