- Introduction
- Table based Questions
- Case1: finding value
- Case2: multiplication chain
- Case3: finding angle
- Mock Questions

# Introduction

For SSC CGL trigonometry, there are mainly four types of questions

- Height and distance: and it has five subtypes. We already saw how to solve them. If you haven’t seen it, click on my youtube playlist
- Table based: Finding values and angles: We will see this in today article.
- Complimentary angles (90-A): discussed in separate article click me.
- Trigonometry formulas + algebraic formulas: to be released.

# Table based Questions

Here is an example of Table based questions:

- What is the value of 3cos
^{2}30+sec^{2}30+2cos-+3sin90-tan^{2}90 - If sinA=2sin30*cos30, then find value of A

- To solve these type of questions, you must know the “table” (=meaning the values of sin, cos, tan, cosesc, sec and cot for 0, 30, 45, 60 and 90 degree angles).
- From the earlier youtube video, you already know the shortcut technique to memorize the values of 30, 45, 60 of sin, cos, tan, cosec, sec, cot, using those two “
*topi triangles*”. (if not click me) - Today we’ll extend it further and create the whole table from 0 to 90 degrees.

0 | 30 | 45 | 60 | 90 | |

Sin | |||||

Cos | |||||

Tan | |||||

Cosec | |||||

Sec | |||||

cot |

First using above “Topi triangles” I’m going to fill up the values of Sin, cos, tan for 30, 45 and 60 degree angles. (sin = opp/hypo, cos=adj/hypo and tan=opp/adj)

but we know know that

- sin = 1/ cosec. That means for cosec 0 to 90, I’ll just inverse the values of sin 0 to 90. (e.g. sin30=1/2 means cosec30 will equal to 2)
- Cos=1/sec. that means for sec 0 to 90, I’ll just inverse the values of cos 0 to 90. (e.g. cos60=1/2 therefore sec60=2).
- Tan=1/cot. That means for cot 0 to 90, I’ll just inverse the values of tan 0 to 90 (e.g. tan45=1, so cot45 will also be 1).

Here is the updated table. (don’t just read this, keep making the table in your notebook/rough paper simultaneously).

Ok so far, we’ve filled up all the values for 30, 45 and 60. Only two column remain: 0 degree and 90 degree. That can be filled up very quickly if you just remember two things

- For zero to ninety, sin is 0 to 1. And the case of cos is reverse (meaning for zero to ninety, cos is 1 to 0).
- Zero’s inverse (1/0)=not defined. And vice versa (meaning “1/not defined” = 0).

- Applying the first rule: for 0 to 90, sin is 0 to 1, and cos is 1 to 0.
- WE also know that Tan=sin/cos.
- So tan0=0/1=0 and
- tan90=1/0=Not defined

- now we’ve the values of sin, cos and tan for 0 degree angle and 90 degree angle. We can inverse them and get the values for cosec, sec and cot also. (recall rule number two: 1/0=not defined and not defined’s inverse is 0).

Finally we’ve our table ready.

I suggest you construct this table on your own for atleast 2-3 times at home. Otherwise this technique will not go in long term memory.

In the exam you don’t have to create the whole table in the rough paper, just get only those values that are required for the given question. Now let’s try some Table based Questions.

# Case1: finding value

**Q. Find value of sin60*sin45+cos60*cos45**

- Just Plug in the values into the given equation
- Answer will be (root6+root2)/4

**Q. if cosA+cosB=2 then find value of sinA+SinB, given that both angle A and B are between 0-90 including both 0 and 90.**

- Ans. If you look at the table, cos0+cos0=1+1=2. That means both A and B angles are 0 degree angles.
- Therefore SinA+SinB=Sin0+sin0=0+0=0.

**Q. Find the value of cos25/sin65. **

- We only know the table for 0, 30,45,60 and 90. We don’t know the tables for 25 or 65. Don’t worry, these questions are based on the formula of complimentary angles. We’ll learn that in next article.

# Case2: multiplication chain

**Q. find the value of tan0 x tan1 x tan2 x tan3x….tan89**

Approach: from the table we know that tan0=0. So no matter what you multiply with zero, final answer will always be zero.

**Q. find the value of cos2 x cos4 x cos6 x cos8x….x cos92**

Approach: as you can see the angles are increasing as per the multiplication table of 2. So in the chain, you’ll also get cos90 (because 2 x 45=90). So we can write the chain as

cos2 x cos4 x cos6 x cos8x….cos88 x **cos90** x cos92

but we know that cos90=0, hence the whole multiplication will become zero.

**Q. find the value of tan48 x tan 23 x tan 42 x tan67**

Approach: as you can see there is no tan0 in this multiplication chain (otherwise we could get zero as the answer.) but this question can be easily solved using the formula of complimentary angles. (we’ll see it in the next article).

# Case3: finding angle

Q. For 0<A<90, if sinA=sin60 x cos30-cos60 x sin30, then what is the value of A?

- 0
- 30
- 45
- 90

Approach:

left hand side LHS | right hand side RHS |

sinA= | sin60 x cos30-cos60 x sin30 |

Plug in the values and you get right hand side is 1/2 | |

sinA= | 1/2 |

Now go back and observe the table. In the Sin’s row, You see the sin30=1/2. Therefore, A=30 degrees.

**Q. If Sin(A+B)=root3/2 and SIN(A-B) =1/2 then what are the values of A and B?** (given that both A and B are acute angles and A>B)

- 30,60
- 45,45
- 45,15
- None of above.

Approach

- Sin(A+B)=root3/2 (this is given in the question itself).
- If you look at the table, sin60=root3/2. That means A+B=60…eq1
- Similarly we’ll get A-B=30….eq2
- So we’ve two equations:
- A+B=60
- A-B=30
- Now add these two equations eq1+eq2
- (A+B)+(A-B)=60+30
- 2A=90
- A=45
- Plug this value back in eq1 (or eq2). And you get B=15
- Final answer C: 45,15

# Mock Questions

## Instructions

- For all these questions, assume that unknown angles are between 0 to 90 (including both 0 and 90).
- Questions about finding values (based on trigonometry table) are easy. You just have to plug in the values and simplify the equation.
- But be sure to observe the rules of simplification (BODMAS) and laws of surds and indices, else you’ll get wrong answer.

- if CosA=1-2sin
^{2}30, then find value of A- 30
- 45
- 60
- 90

- if cosA=2cos
^{2}30-1, then find value of A- 30
- 45
- 60
- 90

- if cos60=cos
^{2}A-sin^{2}30, then find value of A- 60
- 45
- 30
- 90

- if sinA=2sin30*cos30, then find value of A
- 30
- 45
- 60
- 90

- if sinA=2tan30/ (1+tan
^{2}30), then find value of A- 30
- 45
- 60
- 90

- if cosA=(1-tan
^{2}30)/( 1+tan^{2}30), then find value of A- 30
- 45
- 60
- 90

- if cosA=4cos
^{3}30-3cos30, then find value of A- 30
- 45
- 60
- 90

- Find value of (5cos
^{2}60+4sec^{2}30-tan^{2}45)/(sin^{2}30+cos^{2}30)- 67
- 12
- 67/12
- 12/67

- Find value of 3cos
^{2}30+sec^{2}30+2cos0+3sin90-tan^{2}60- 67
- 12
- 67/12
- 12/67

- If sinA=cosA, what is the value of 2tan
^{2}A+sin^{2}A+1- 2
- 7
- 7/2
- 2/7

- What is the value of sinA*cosB+cosA*sinB, if A=30 and B=60
- 0
- 1/2
- 2
- 1

- What is the value of cosAcosB-sinAsinB, if A=30 and B=60
- 0
- 1/2
- 2
- 1

- secA=cosec60, what is the value of 2cos
^{2}A-1- 0
- 1/2
- 2
- 1

- What is the value of 5cos
^{2}90+3sec^{2}30+4cos^{2}45+tan^{2}60- 3
- 7
- 5
- 9

- What is the value of 5cos90-cot30+(sin60/cos
^{2}45)- 1
- 3
- 0
- 5

- Find the value of (cos30+sin60)/(sin30+cos60+1)
- 1/2
- 1/root2
- Root3/2
- 2/root3

- Find the value of tan
^{2}60/(sin^{2}45+cos^{2}45)- 3
- 1/2
- 2/3
- 1/3

- If cot(A+B)=1/root3 and Cot(A-B)=root3. Find the values of A and B
- 30,60
- 60,30
- 15,45
- 45,15

- Find the value of cos10 x cos20 x cos30 x….x cos90
- 1
- 1/2
- 0
- Not defined.

- If secA-cosecA=0 then find value of secA+cosecA, given that A is an acute angle.
- Root 2
- 2 root 2
- 0
- Not defined

# Answers

**1)c, 2)c, 3)c, 4)c, 5)c, 6)c, 7)d, 8)c, 9)c, 10)c, 11)d, 12)a, 13)b, 14)d, 15)c, 16)c, 17)a, 18)d, 19)c, 20)b**

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

## So far 63 Comments posted

good but very basic fundas…good for beginners.

FINALLY…..I WAS WAITING FOR THIS ARTICLE FOR SO LONG AND I HAVE MY EXAM NEXT WEEK.THANX MRUNAL.IF YOU CAN PLEASE UPLOAD AN ARTICLE ON DATA INTERPRETATION MAINLY RELATED TO PIE CHARTS ASAP.

Thanks Sir. It is really very helpful.

Sir please upload some helpful articles for Mensuration as well. And sir,if I complete only NCERT for this exam,then it would be helpful kya? Or I should do some practice from Sarvesh Verma book also?

Mrunal ,

Question , tan 90 is ND and anything added/subtracted to ND = ND.

Am i wrong somewhere or any typo ?

Tan^2 60 will give the exact answer.

sorry , Question 9.

9th question. sorry typing mistake on my part. it was tan

^{2}60 (and not tan^{2}90)14.ans 5+2root3

15.ans root3/2

16. Ans 2root3/3+root3

19. Degree is not mentioned… so it is radian 1 radian=57.3degree

Sir one doubt Please clarify me

For Q.No 9) Value-ND (I was getting)

Ans is ND or Value(103/12)

Q.No 19) ans is zero (bcoz cos90=0)

For Q9 Ans will be ND only as tan90 is ND & any operation with ND results in ND only.

Mrunal Sir, Please clarify.

Hi Arun I am also getting the same answer for Q.No 14,15,16

For Question 19 correct answer should be C as Cos 90 is 0

Am I eligible for Staff Selection Commission Recruitment 2013 in Cabinet Secretariat? I have done my graduation in English. Please help. IN educational qualification it says “Research Officer : Honours degree / PG in Economics / Commerce.” This is pretty confusing. Help needed.

you are eligible only for ground duty

Thanx a lot mrunalbhai…As usual commendable work…

sir, draft policy on PPP has been put up on finance min. site… it provides usefull info such as defination of PPP’s and its mode… can u streamline the document and point out imp things in it in ur future article… will be greatful

Hi can any one tell me how to get answer for Question 10 🙂

actually there was a typing mistake.

The question 10 is if sinA=CosA….then (so A=45 degrees, so you apply the values in the equation accordingly)

Thanks Bhaiyya :):)

sorry for the Typing errors in the mock questionsHere are the corrections:

Q.9 => last part was tan60 and not tan90

Q.10=> it was sinA=cosB.

Q.14=> 5cos

^{2}90+3sec+4cos^{2}30^{2}45+tan^{2}60Q.15=>5cos90-

cot30+(sin60/cos245)Q.16=>(cos30+sin60)/(

sin30+cos60+1)Q.19=>answer was C (zero)

Dear Sir,

This is my first interaction with you. I dont know how to tell you but any how i want to get out of it. so here is my problem, sir i have done 12th, Diploma & BE in civil engg. Now my age is 28 running. I want to do Civil cervices, but the problem is now i m doing PGP in Quantity Surveying And in search of good private job. but my interim wish is to do civil services, but the situation becomes very crtical for me. because my family wants me to do private job.They dont have faith on me that i can do this civil services. so now i m having double mind situation. I m a average student in studies, and unable to generate confidence to do civil cervices. I am very poor in planning the things. life is going so confusing stage. No one is there for guide lines. Always i have decided to do so many things but always unable to implement it.There are so many things that i have to tell but cant express myself properly. so sir what can i do now? how can i improve myself?

read this:

http://mrunal.org/2012/05/topper-om-air17-cse2011.html#140

thanks sir

Dear

Hi!

I am getting an issue in respect of medical exam for non defence posts of CGLE 2013.

I have filled up the form this year and hope will make it even. But, I have some internal sex organ issue in my testes.My one testes is normal but other one is very small due to an injury in childhood.

So, I want to know that whether I can get a job in non defence posts or SSC as ITO or assistant etc or not.?

Please guide me to get out of this issue.

I will be thankful to you

Sorryyyy

TABLE ARE NOT SHOWNNN

PLS HELPPP

try with a different browser, make sure images are not disabled.

Thanks a lot of for you great help for the students

Hey,could someone explain question 7,I keep getting 3root3-3root3/2

Cos A = 4 Cos cube 30-3 Cos 30

= 4(root3/2)cube- 3(root3/2)

= root3 /2 – root3 /2=0

4cos^3(30) is also 3root3/2…you even dnt need to calculate..thats COS3A formula..implies angle=3*30=90 degree

🙂

Hi Sir,

Could you please help me out.

I got SBI PO written exam call letter notification on 06-Apr-2013 and I checked today(08-Apr-2013)

in bank website but it showing “Last date to download call letter over”.

Is there anyway to get my call letter for written exam which is to be held on 28-Apr-2013?

Same thing happened to me. Why has sbi closed its website? That’s not fair. Now what we r gonna do

I have downloaded my call letter today only. They have uploaded on their site. Check again on sbi.co.in.

Admit card is available online from 08-04-2013.I have downloaded it right now

this is the link to download SBI PO call letters: http://ibpsreg.sifyitest.com/probsbiapr13/login.php

@prakash@ No need to worry they have dowloaded the previous year admit card by mistake, the time they realised they took it out frm website

Mrunal Sir, Please solve q no 20 and this one also – if tan(#1 + #2)= root 3 and sec(#1-#2)= 2/root3

then find the value of sin 2#1+tan3#2

# means theta…

Hi aditya,

qn20 secA=cosec A when A=45 (see table above)deg=> ans is root 2 +root2=2 root2

Asso

#1+#2=60deg,#1-#2=30deg(see above table)=> #1=45deg,#2=15deg 🙂

sin 90+tan45=1+1=2

can any one explain question no 10..

If sinA=cosA, what is the value of 2tan2A+sin2A+1

sinA=cosA

sinA/cosA=1

tanA=1

A=45

given 2tan^2A+sin^2A+1

2(tan^2 45)+sin^2 45+ 1

2(1)+(1/2)+1

2+1/2+1

3+1/2

7/2

required ans=7/2..

Thanks Mrunalsir

Thank you very much sir..

Thank you so much sir for making geometry and trigonometry easy….i am happy now

Dear

Hi!

I am getting an issue in respect of medical exam for non defense posts of SSC-CGL 2013.

my brother has only one testes by birth he has filled up the form this year and hope will make it even.

So, I want to know on behalf of my him as he is little bit disturbed due to this that whether he can get a job in non defense posts or SSC as ITO or assistant etc or not.?

Please guide me so that i can meke him get out of this issue.

I will be thankful to you.

please reply sir

This is a awsome helpful.site for ssc-cgl aspirants. Thank u very much.

another method for making sine table:

sin0 = square root of 0 / square root of 4

sin 30 = square root of 1/ square root of 4

sin 45 = square root of 2/ square root of 4

sin 60 =square root of 3/ square root of 4

sin 90 = square root of 4/ square root of 4

make the table and do it reverse for cos

can anyone explain 18 20

Are these 5 basic types enough for ssc cgl ? Please tell

Mrunal you are GOD for students

Mrunal you are GOD for students

Mrunal you are GOD for students

Mrunal you are GOD for students

Mrunal you are GOD for students

pehle app ko main apne bare me bta du.. I just join tcs.. i am 23 now.. I have 68.8 in 10th and 74.6 in 12th and 65.8 in BE (information technology ) …So what are the option open for me.. you suggest me some thing And I am GEM candidate… i have lots of extra curricular but dont of any importance ….So plz suggest me sum thing related to my profile … and ya i wanna be ias…

What are the options left for me… i am realyy tensed and wanna …suicide..

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

This is an awsome & helpful.site for all competative exams aspirants. Thank u very much.

GREAT JOB. . . . . . MRUNAL SIR

plz include few SSC type tricky problems as well.

nice

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

question 7…….mine answer 4…

solution plzz??

if cosA=4cos^3(30)-3cos30, then find value of A

cos30=Root 2

so 4*3Root 3/8-3*Root3

3 root 3/2-3 root 3/2=0

cos A is zero if angle is 90

so ander is option d

if cosA=4cos^3(30)-3cos30, then find value of A

cos30=Root 3/2

so 4*3Root 3/8-3*Root3/2

3 root 3/2-3 root 3/2=0

cos A is zero if angle is 90

so ander is option d

I am really thankful to the author and all the guys who had put up their valuable time while explaining such basic but important concepts. Finally I could do these type of questions without much difficulty. Values can’t be forgotten at all now all thanks to the “Topi Triangles” method. Wish my maths’ teacher were this good in explaining these concepts. Thanks a lot again guys. Please keep up with the good work.

Woww! Sir ur explanations in quantitative aptitude are so simple and these are very usefull and in so simplyfied language . I’m very pleased to have this great gadget and teacher on my device ..

thank u sir …u are helping us lot …

if atriangle ,if b+c divide 11=c+a divide 12=a+b divide 13 prove that cosA/4=cosB/19=cosc/25?

can any1 help me with question 13???