- 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: ** viagra generico 25 mg italia pagamento online a Parma sinA=cos(90-A)** and ** here CosA=sin(90-A). **

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

Le’s try a question

** dove comprare viagra generico 25 mg a Milano 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

** cialis generico miglior prezzo 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 ** go to site 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.

** acquistare viagra generico 100 mg a Torino 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)

** http://buy-generic-clomid.com 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

** go here 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

## 61 Comments on “[Trigonometry] Type#3: Complimentary Angles related Questions and approach for SSC CGL”

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 .