- What is complimentary angles?
- SIN vs COS
- SEC vs COSEC
- TAN vs COT
- Multiplication chain (tan and cot)
- Finding unknown angle
- The half angles
- Summary
- Mock question

# What is complimentary angles?

- If you add two angles (A+B) and if their sum is 90, then they’re called complimentary angles.
- For example, 60 and 30 are complimentary because 60+30=90.
- Same way (0,90)
- Same way (45,45)
- Same way (85,5), (12,78), (15,75) and so on…
- If A and B are complimentary then A+B=90. Based on this we can also say that,
- B=90-A and
- A=90-B.

# SIN vs COS

In the earlier article, we saw how to construct and memorize the trigonometry table for sin, cos, tan, cosec, sec and cot for 0, 30, 45, 60 and 90 degrees. Click me if you did not read it

If you compare the rows of sin and cos. You’ll notice that

- Sin0=cos90=0
- Sin30=cos60=1/2
- Sin45=cos45=1/root2
- Sin60=cos30=root3/2

We can generalize this as: **sinA=cos(90-A)** and **CosA=sin(90-A). **

This generalization is true for all angles between 0 to 90 including 0 and 90.

Le’s try a question

**Q. Find the value of cos18/sin72**

## Approach

- We know that cos and sin are complimentary.
- CosA=sin(90-A)
- So for Cos18, I can write cos18=sin(90-18)=sin72. Let’s use it

**Cos18/sin72**

=sin72/sin72 (because cos18=sin72).

=1 (because numerator and denominator are same so they’ll cancel eachother.)

## Food for thought

- You’ve to convert only one of the two values. For example, in above sum we converted Cos into its complimentary Sin. You could also solve it by converting the sin into its complimentary cos. That is sin72=cos(90-72)=cos18. And then cos18/sin72=cos18/cos18=1.
- If you convert both values then you’ll never get answer. (that is cos18/sin72=> converted to sin72/cos18). In that case you’ll run into an infinite loop!
- In the exam, sin and cos’s complimentary concept will help mostly for divisions and substraction cases.

Division cases | Substation cases |

Cos18/sin72=sin72/sin72=1. | Cos18-sin72=sin72-sin72=0 |

Here they numerator and denominator cancels each other. | Here too they’ll cancel each other. |

But suppose if there is cos18+sin72, this doesn’t lead to any value. (because

In the exam, sin and cos’s complimentary concept usually **doesnot help in multiplication or addition cases.**

## Multiplication cases | ## Addition cases |

Cos18 x sin72 = sin72 x sin72 = sin^{2}72. But this is dead end because we don’t know the value of sin72 | Cos18 + sin72=sin72 + sin72= 2sin72. Again dead end because we don’t know the value of sin72. |

Point being: in the exam, after doing some steps, if you run into above situations, there is a chance that you’ve made some mistake in calculation or you’ve moved into the wrong direction. Although there can be exceptional situation where it could help, when they’ve framed equation using both trigonometry + algebraic equations (That is Type#4 questions, we’ll see about it in next article).

# SEC vs COSEC

Just like sin and cos, we can see that sec and cosec are also complimentary. (in the table you can see that for (30,60) and (45,45) the values of sec and cosec are same. Therefore,

- secA=cosec(90-A). but this is not valid for 90 degrees because sec90 is not defined.
- cosecA=sec(90-A). but this is not valid for 0 degree because cosec0 is not defined.

**Q. find value of sec70 x sin20 – cos20 x cosec70**

If you convert all four (sin, cos, cosec, and sec) into their complimentaries (cos, sin, sec and cosec) then you’ll run into infinite loop.

Part A minus | Part b |

Sec70 x sin20- | Cos20 x cosec70 |

In the part A, if I convert sin into its complimentary cos, then sec x cos =1 because sec and cos are inverse of each other. Same way in part B if I convert cosec into its complimentary sec, that too will lead to 1. Let’s see

Part A – | Part b |

Sec70 x cos(90-20)- | Cos20 x cosec(90-70) |

Sec70 x cos70- | Cos20 x sec20 |

1- | 1 |

=1-1

=0 is the final answer.

# TAN vs COT

In the table, you can see that

- Tan30=cot60=1/root3
- Tan45=cot45=1
- Tan60=cot30=root.

So, tan and cot are complimentary

- tanA=cot(90-A) but not valid for 90 degrees because tan90 is not defined.
- cotA=tan(90-A) but this is not valid for 0 degree because cot0 is not defined.

The tan and cot’s complimentary relationship is also import for multiplication chain questions because tan and cot are also inverse of each other. That is tan A x cot A= tan A x 1/tan A=1.

# Multiplication chain (tan and cot)

**Q. Find value of tan48 x tan23 x tan42 x tan67**

When you get this type multiplication chain question, you’ve to find out the pair of complimentary angles. Here 48 + 42=90 and 23 +67=90 so I’ll club them together

Part A x | Part B |

Tan48 x tan 42x | Tan23 x tan67 |

In both parts, we’ll convert any one tan into cot, then tan x cot = 1 (because they’re inverse of each other).

Part A x | Part B |

Tan48 x cot (90-42)x | Tan23 x cot(90-67) |

=tan48 x cot48 | Tan23 x cot23 |

=1x | 1 |

Final answer =1.

Let’s try another one

**Q. Find value of tan1 x tan 2 x tan3 x ….x…tan88 x tan89**

Approach: you make pairs of complimentary numbers: 1+89=90, 2+88=90…. There is only one angle left who doesn’t get a pair (45)

So it’ll look like this

= (tan1 x ta89) (tan2 x tan88)x…xtan45

In each of those pairs, you convert one tan into its complimentary cot.

=(tan1 x tan(90-89)) x (tan2 x cot(90-88))…..x1 ; because tan45=1

=(tan1xcot1) x (tan2xcot2)x…..x1

=1 because tan and cot are inverse of each other.

# Finding unknown angle

Q. given that 4A is an acute angle and sec4A=cosec(A-20), what is the value of A?

Ans.

LHS | RHS |

Sec4A | Cosec(A-20)=sec(90-(A-20)) =sec(90-A+20) |

Sec4A | =sec(110-A) |

Since LHS=RHS so we can say that on both sides the angles are equal

- 4A=110-A
- 5A=110
- A=110/5=22 degrees.

# The half angles

Q. in triangle ABC, if A, B and C are acute angles, then Cos(B+C)/2=sin(A/2).

- This statement is always correct
- This statement is always Incorrect
- Depends on values of A,B and C
- None of above.

## Approach

We know that in any triangle ABC, the sum of three angle A+B+C=always 180

- B+C=180-A ; (I’m taking c on right side)
- (B+C)/2=(180-A)/2 ;(I’m dividing both sides by 2)
- (B+C)/2=(180/2 MINUS A/2)
- (B+C)/2=[90-(A/2)]……let’s call this equation 1.

Now back to our complimentary formulas. we already know that for angles between 0 to 90, sinX=cos(90-x) this is definitely correct. So instead of “x”, I’ll now write A/2

Sin(A/2)=cos[90-(A/2)]…..eq2

But according to equation 1, [90-(A/2)]= (B+C)/2

Putting that back in eq2

Sin(A/2)=cos(B+C)/2. Therefore answer is (A) always correct.

# Summary

COMPLIMENTARY(90-A) | INVERSE (1/A) | |

SIN | COS | COSEC |

COS | SIN | SEC |

TAN | COT | COT |

COSEC | SEC | SIN |

SEC | COSEC | COS |

COT | TAN | TAN |

These complimentary are valid for acute angles (between 0 to 90) and some of them are not valid for 0 or 90 if they’re “not defined”.

- But what about angles greater than 90 degree. For example, sin150? For that we’ve to look at the quadrants and change the signs accordingly. (click me for more) While that concept is important for CAT, but as far as SSC is concerned, there is not much point into going that deep.
- In the SSC exam, if you’re given a lengthy trigonometry equation and asked to find its value or angle then what to do? Well three situations can happen:

- If you’re given an equation that contains 0, 30, 45, 60 and or 90 angles of sin/cos/tan/cosec/sec or cot then you just have to plug in the values from trig.table and simplify the equation. You’ll get the answer. (the complimentary case usually doesn’t apply in such situations).
- If you’re given an equation that contains odd looking angles such as 12,15, 43, 85 or anything that is not from the (0/30/45/60/90) group, then you’ve to think about the application of complimentary angles.
- You’ve to make “pairs” of complimentary angles (e.g. sin10/cos80) and then convert ANY ONE in the given pair. If you convert both into their complimentary, then you’ll never get the answer.
- Complimentary pair (sin|cos), (cosec|sec), usually leads to division or subtraction then most of the stuff in the equation gets eliminated and you get “good looking number” as answer for example 1,2,3,4,1/2, 1/root3 etc. (if such pair leads to multiplication case then perhaps their inverse is present (sin x cosec) or (cos x sec).
- Complimentary pair (tan|cot), usually leads to multiplication. Because tan and cot are also inverse of each other (tan x cot = 1), and then most of the things in the equations get eliminated. But keep in mind tan and cot are inverse only for same angle. E.g. tan33 x cot33=1. But tan33xcot34=not equal to 1.

- If above situation 1 and 2 doesn’t help, then it is most likely a question involving trigonometry formulas like
**sin**, or^{2}A+cos^{2}A=1**sec**or^{2}A-tan^{2}A=1- cosec
^{2}A-cot^{2}A=1 (or all three of them). - and or the Application of algebraic formulas such as
**x**^{2}-y^{2}=(x-y)(x+y). - we’ll see about such scenarios in the next article.

# Mock question

- Find value of cosec32-sec58
- 1
- 2
- 3
- 0

- Find value of Cot12 x cot38 x cot52 x cot60 x cot78
- 1
- 0
- Root3
- 1/root3

- Root3 x (tan10 x tan30 x tan40 x tan50 x tan80)
- 0
- 1
- 3
- 2

- Cos48 x cosec42 + sin48 x sec42
- 0
- 1
- 2
- 3

- (cos70/sin20)+(cos59 x cosec31)
- 1
- 2
- 3
- 0

- (sec20/cosec70)+[(
**cos55 x cosec35**)/(*tan5 x tan25 x tan45 x tan65 x tan85*)]- 1
- 2
- 3
- 0

- (sin70/cos20)+(cosec20/sec70)-(2cos70 x cosec20)
- 1
- 2
- 3
- 0

- If sin3A=cos(A-26), then what is the value of A? given that 3A is an acute angle.
- 21
- 23
- 24
- 29

- If sec2A=cosec(A-42), then what is the value of A? given that 2A is an acute angle.
- 22
- 33
- 44
- 55

- If sinA=cos30, what is the value of
**2tan**, given that A is an acute angle.^{2}A-tan45- 3
- 4
- 5
- 7

- If secA=cosec60, what is the value of
**2cos**, given that A is an acute angle?^{2}A-1- 0
- 1/2
- 1
- 2

- If cot2A=tan(A+6), find the value of A. given that 2A and A+6 are acute angles
- 18
- 28
- 30
- None of above.

- What is the value of cosec(40+A)+tan(55+A)-Sec(50-A)-cot(35-A)
- 1
- 2
- 3
- 0

- For triangle ABC, which of the following formula is incorrect? (this was asked in SSC)
- COS (a+b)/2= SIN c/2
- SIN (a+b)/2= COS c/2
- COT (a+b)/2=TAN c/2
- TAN (a+b)/2=SEC c/2

## Answer

1)d, 2)d, 3)b, 4)c, 5)b, 6)b, 7)d, 8)d, 9)c, 10)c, 11)b, 12)b,13)d, 14).d because TAN and SEC are not complimentary.

For more articles on trigonometry and aptitude, visit Mrunal.org/aptitude

## So far 61 Comments posted

Hi, Brother i have lost my Registaration ID of SSC CGL, please help how to regaing it from SSC Website, or by any other mean. please help Thanks

Hey,

Just enter your name and date of birth to download the admit card.

http://sscnr.net.in/newlook/Admitcard_CGLE2013/CheckRoll.aspx

its showing ‘Sorry you have entered wrong Roll Number or Date Of Birth’ , even on entering correct details, i guess its thing to worry, might be my form is not accepted, but at the end of filling part 2 of form, it showed successful!!

Check out the list of incomplete forms on http://www.ssc.nic.in

you can get your roll num. By putting ur name and dob . .Ssc site do have that option also

No need of reg. id.You can download ur admit card from ur regional ssc website, it requires only ur name and DOB (date of birth)

can you give the ssc website link i am not getting it

i am from North western region, punjab chandigarh, on website its asking ‘enter First 3 character’ of name and father’s name, but when i enter its showing some other person admit cart, there must be many with same first 3 charater name and father’s name!! what to do now.

i have written mail now to the email address given on website, is there any ther alternavive, is yes, please tell anybody

im from chandigarh also

goto sscnwr.org

click red colored DUPLICATE ADMIT CARD link

enter 1st 3 letters or

1. Ur name

2. Ur father’s name

ull get a list of candidates, select yours by verifying DOB..

If you know/remember your name and Date of Birth then u can download your Admit Card from the respective sites of SSC . For Northern region it is http://sscnr.net.in/newlook/Admitcard_CGLE2013/CheckRoll.aspx

Yes Brother, i do remember my name and DOB 🙂

Thanks

sir ji,

trigno mein max. Or min. Value nikalne wala quest. Har baar aata hai. Aaj tak kisi ne bi wo solve karne ki approach ni btai. Aap bta dijiye.

For eg.

q1. Min. Value of 6 Sin^2 (x) + 9 cos ^2(x)

q2. Max. Value of 4sec^2 (x) + 5 tan^2 (x)

q3. Sin 2x ,cos 4x ,tan 2x , sec 4x , cosec^2 (x) and sin^2(x) mein alag alag max. And min. Value kya hogi

thanku

Hello DS,

Min value of Sin and cos functions can be between ( -1 , to , 1) . (that is Range = (-1 ,1), for any domain.. Tum bas itna yaad rakho ki Sin(x) aur Cos(x) , ki Minimum value ho sakti ha = -1 and Maximum value ho sakti ha = 1.

Ok , now back to your questions,

Question 1) 6 Sin^2 (x) + 9 cos^2(x)

= 6 ( sin^2 (x)+ Cos^2 (x) ) + 3 Cos^2 (x)……..1

Now we know that Sin^2 (x) + cos^2(x) = 1 , put this value in eqn 1, we get,

6 ( 1 ) + 3 Cos^2 (x),

Now , when Cos(X) , will be Maximum then Cos(x) = 1 , thus

Maximum value of eqn-2, will occur when Cos(x)= 1 , thus

Maximum Value = 6( 1) + 3 * 1

= 9

Similarly Minimum value occur at Cos(x) = -1 , thus

Minimum value = 6 – 3 = 3.

Next question is similar , do on your own and use identity ( 1+ tan sq(x) = Sec sq(x).)

Now here is one excercise for you :- by converting all trigonometric ratios into sin and cos , find Max and Min value of all Trigonometric ratios. And that will be the answer to your Third question.

min value =n 1 ki 3 kaise?

cos min is -1 max is 1

cos^2 min is 0, max is 1, fir 6 + 3(cos^2(180)) = 9

and if angle is 90 , then 6 + 3 (cos^2(90)) = 6 + 3(0) = 6 is minimum..

Hi HK, yaar I am not much accustomed to see square on computer and also I posted a very lengthy comment , so while I was typing It just slipped out of my mind that Cos square (2x) is given in question , I just misread that while typing.

Yes the minimum value of Cos SQ(2x) occur at x =0 and thus will be 6 . thank you for notifying. and sorry that I made that mistake.

Sorry its at cosx = 0 , that is X = 2n *Pi+Pi/2 or 2n* Pi + 90 or at 90 degrees.

yaar type krne me error ho jata ha.

Dear Fanish ji

As per ur explanation for the minimum value

6 + 3 cos^2(x) if u put cos(x) = -1 thn again Min value comes out to be 9 (boss square hai)

in this case put cos(x) = 0

then u will get min value = 6

Hi HK, yaar I am not much accustomed to see square on computer and also I posted a very lengthy comment , so while I was typing It just slipped out of my mind that Cos square (2x) is given in question , I just misread that while typing.

Yes the minimum value of Cos SQ(2x) occur at x =0 and thus will be 6 . thank you for notifying. and sorry that I made that mistake.

Sorry its at cosx = 0 , that is X = 2n *Pi+Pi/2 or 2n* Pi + 90 or at 90 degrees.

yaar type krne me error ho jata ha.

hello fanish,

ques. Is

2sin^x + 3cos^x

nw

after breaking and simplyfying we get

2 + cos^x. .putting cos^x min. Value 0 . .we get 2 as min. Value. .

Nw just convert cos^x to sin x we get

2sin^x + 3(1-sin^x). .simplifying we get it as

3-sin^x

nw put sinx min. Value we get 3 as min. Value. .

But both can’t be min. Value .. Then what to do in such cases ? ?

Hi…a simple suggestion..don’t take algebra here…sin x(these are functions not a simple quadratic equation) curve is difference ,sin^2 x,sin^3x are different…the max and min values are different

For Sin x ,[-1,1] for sin 2 x its [0,1]..its easy if u can imagine a curve where valule for angle between 180 and 360 is also +ve

For above problem convert into 1 function you may get

(1)f(x)=9-3sin^2(x)

since there’s -ve sign in between,

(a)f(x) will have min when sin^2(x) is max=)9-3(1)=6

(b)f(x) will have max when sin^2(x) is min=)9-3(0)=9

(2)6+3Cos^2(x)

(a)f(x) will have min when cos^2(x) is min=)6+3(0)=6

(b)f(x) will have max when cos^2(x) is max=)6+3(1)=9

You can even apply maxima and minima concept of single derivatives and double derivatives ,but not necessary for this exam..

Tell me if i’m wrong anywhere guys..

sin^2(x) and (sin x)^2..they both are entirely different

Sorry guys..a little bit correction..sin^2(x) and (sinx)^2 are same only..its only when there are -ve exponents,they bring different meaning ..say (sin^-1)(x) and (sinx)^-1..

Sorry Minimum value occur at cosx = 0 , that is X = 2n *Pi+Pi/2 or 2n* Pi + 90 or at 90 degrees.

It just slipped out of my attention that Square of function was given in the question.

sure very soon

sir how to solve max. Min. Type of ques. ?

Your original Admit card might be on the way to your Correspondence address,so there is no need to panic as such! If you are facing such problem(Duplicate Admit Card) you should contact at the number in the site sscnwr.org rather than e-mailing!

Also,for once you enter 3 letters each of your name and your father’s name,there’s a list generated of the people with the similar names,if any,you have to spot out yours and select yours and get the Print out of your respective admit card!

Try again!Hope it helps!Best of luck!

sscnwr.org did not generate any list as such, its showing only a single person with that particular detail, that’s why i asked. If it would have generated a list, it would not have been a problem. and thanks the info if its delivered to Correspondence address, its good!! and i will try to call if its not delivered within few days, exam date is 21st, so i’ll be calling in 10 days befor exam, i think that would do.

Thanks!

hello sir, can u plz explain how to solve crptarthim…..i have to give elitmus test and there are 4-5 questions based on crypyarthim…

Sin and cos lies bw [-1,1]

Cosec and sec lies (-infinity,-1]and [1,infinity]

Tan tan and cot lies bw (-infinity, +infinity)

Sir, sometimes questions are asked to find minimum or maximum value of some expression(generally trigonometrical), how to find that?

Thank u sir…

Sir…Any idea about CSIR answer Key

hi,

Plz let me know the formula for questions like :

cos^n x + Sin ^n x =???

Hey frnds me too facin same prob like rahul pls help me out n more over

i have got a confirmation mail from ssc, my exams are on 14th….

pls give the phone number of ssc.

Hey frnds i got my admit card,rahul try typin ur name if u have any initial then dont put full stop in between ur name n initial just leave space….i was actually goin wrong here try doin n u ll surely get it.

Sir plz write next article (trigno) as soon as possible… coz my exam is on 14th…

Guys if i enter my Resistration ID in the SSC website for downloading Hall ticket,it’s showing no such number found.Is any body facing the problem.Please help me out(i have registerd successfully)

read my comment above hope u might have facin the same….

sandeep pls read my comment above hope the same u might have facin….

plz anyone explain me answers for 8,9,10 questions

Thanks Mrunalsir

where will i get the solutions of your mock test not answer proper solutions esp. from q 11 to 14

btw nice exercise i was able to solve enough by overcoming maths phobia

The angles of elevation of the top of a tower standing on a plane from two points on a line passing through the foot of the tower at a distance 9 ft and 16 ft respectively are complementary angels. then the height of the tower is?

Sir,plz tell me how to approach such problems?

let’s assume that the point that is 9 ft away from the tower makes an angle of “A” degrees.

and point that is 16ft away from tower makes an angle of “90-A” degrees.

The height of the tower is same in each case (d ft).

tanA=d/9……eq1

Tan(90-a)=d/16

=>cotA=d/16….eq2

multiply both equations:

tanA x cot A= d/9 x d/16

but tan x cot =1

=>1=d/9 x d/16

=>d^2=9 x 16=(3)^2(4)^2

=>d=3 x 4=12 ft.

sir,

is lic ado result decleared ?

yes ADO written exam result was declared few days ago.

tan 1 X tan 2 …….. tan 89 wala Q to as it is aa gya boss… tussi great ho sir ji….

Pdf of tier 2 books wanted. esp. Apti… Help me guys.

can any one explain me 12 13

Thnx sir for aur valuable knowledge.

Thnx sir for your valuable knowledge.

Mrunal Sir , i want to ask you that whether the theories and problems of Trignometry given by you is sufficient for solving question related to Trignometry in SSC tier 1, i mean will i be able to solve trignometry parts by studying only article given by you …Please reply

please solve anyone. its answer is 3/4 <= A <= 1

if cos θ+ sin θ = √2 cos θ then cos θ – sin θ = ? can anybody pls solve

whenever this type of expression comes use the (a+b)2 expression or some identities related to that

squaring the question 1+2sin$cos$=2cos2$ and (cos$-sin$)2 =1+1-2cos2$=2sin2$ and ans is obtained

and for the minima and maxima of tringni use the gm am approach when the expressions are cotangent values…………

sin(90°_3a)cosec42°=1 proove it

mrunal sir , i need the current affair for ssc examination . i am in poor in mathematics and english . could you suggest me what are the books are eaisly to prepare ? plz help me .