- Algebra: 3 formulas
- Trigonometry: Formulas
- Case: Only trig formulas
- Case: Combo of Trig + algebra formulas
- Mock Questions
Before moving to the next topic of trigonometry, I would like to quote the tweets of SSC chairman’s official account:
What SSC chief said
- In SSC CGL 2013, the Maths will be of class 10th level.(tweet link)
- …questions in trigonometry and geometry will certainly be easier.(tweet link)
- (because)… some non mathematics candidates (in 12th std) had complained about trigonometry and geometry questions of Higher Secondary level (class12) level (in earlier SSC CGL exam).(tweet link)
Point being: there is no need to be afraid of either trigonometry or geometry. You need to understand a few concepts, mugup a few formulas and practice questions, then both trigonometry and geometry can be solved without much problem. Anyways, back to the trig. Topics. Until now we saw
- Height and distance problems
- How to construct the trigonometry table and solve questions
- Complimentary angles (90-A)
- now moving to the last major topic: the questions based on combo of Trig+algebra formulas. Almost all of them can be solved by mugging up only six formulas: 3 from algebra and 3 from trigonometry.
Algebra: 3 formulas
These three formulas are
- (A plus or minus B)2
- A2-B2
- (A plus or minus B)3
Let’s see
First formula
| Positive sign | Negative sign |
| (a+b)2=a2+2ab+b2 | (a-b)2=a2-2ab+b2 |
Second formula
(a2-b2)=(a+b)(a-b)
Third formula
| Positive sign | Negative |
| a3+b3=(a+b)3-3ab(a+b) | a3-b3=(a-b)3+3ab(a-b) |
based on above formulas, we can derive some more formulas for example
(a+b)2=a2+2ab+b2=> a2+b2=(a+b)2-2ab
Same way,
a3+b3=(a+b)3-3ab(a+b)
=> (a+b)3= a3+b3+3ab(a+b)= a3+b3+3ab2+3a2b
Trigonometry: Formulas
Again, only three formulas
- Sin2a+cos2a=1
- Sec2a-tan2a=1
- Cosec2a-cot2a=1
you can verify these formulas by plugging the values from our trigonometry table.
It is important not to make mistake in these three fomulas vs complimentary angle
| Complimentary angle (90-A) | Trig.formulas |
| Sin and Cos | Here also both are together. Sin and cos: Sin2a+cos2a=1 |
| Cosec and sec | Here cosec comes with cot (and not Sec). Cosec2a-cot2a=1 |
| Tan and cot | Here Tan comes with sec (and not Cot): Sec2a-tan2a=1 |
Based on these formula, you can derive many other formulas for example
Tan=sin/cos.
And we now know that Sin2a+cos2a=1. So
cos2A=1-sin2a
cosA=sq.root of (1-sin2a)
now pluggin this back into Tan’s ratio
TanA=sinA/cosA
=sinA/ sq.root(1-sin2A)
In some books you might have seen such complex formulas for cosec, sec, cot also. Basically they’re derived from the three main formulas given above..
Case: Only trig formulas
Q. Find the value of sin2a+[1/(1+tan2a)]
We know the Sec2a-tan2a=1=>sec2a=1+tan2a
Substituting this in the question
sin2a+1/(1+tan2a)
=sin2a+(1/sec2a) ; but cos= 1/sec
= Sin2a+cos2a
=1
Q. if 3cos2A+7sin2A=4 then find value of cotA, given that A is an acute angle?
Approach
Instead of 7sin2a=3sin2a+4sin2a (because 4+3=7)….eq1
Now let’s look at the question
3cos2a+7sin2a=4
3cos2a+3sin2a+4sin2a =4 ; using value from eq1
3(cos2a+sin2a) +4sin2a =4
3(1)+ 4sin2a =4
4sin2a=4-3
Sin2A=1/4
sinA=squareroot (1/4)=1/2
From the trigonometry table (or Topi Triangles), we know that sin30=1/2 therefore angle “A” has to be 30 degrees. And from the same trigonometry table, we can see that cot30=root3. That’s our final answer.
Q. if cos2a+cos4a=1, what is the value of tan2a+tan4a
Approach
We know that Sin2a+cos2a=1 =>sin2a=1-cos2a…eq1
In the question, we are given that
cos2a+cos4a=1
taking cos2a on the right hand side
cos4a=1-cos2a
taking value from eq1
cos4a=sin2a
cos2a x cos2a = sin2a ; laws of surds and indices
cos2a=(sin2a/cos2a)
cos2a=tan2a……eq2
now lets move to the next part. in the question, we have to find the value of
tan2a+tan4a
=tan2a+(tan2a)2 ; because (a2)2=a2×2=4
= cos2a+cos4a ;applying value from eq2
=1 ;this was already given in the first part of the question itself.
Final answer=1.
Case: Combo of Trig + algebra formulas
Q.find value of sin4a-cos4a
We know that
(a2-b2)=(a+b)(a-b)
So instead of sin4a-cos4a, I can write
(sin2a)2-(cos2a)2 ; because a4=2×2 =(a2)2
=(Sin2a+cos2a)(Sin2a-cos2a)
=1 x (Sin2a-cos2a) ; because (a2-b2)=(a+b)(a-b)
=(Sin2a-cos2a)
We can further simplify this as (sina-cosa)x(sina-cosb)
Q. if sin4a-cos4a=-2/3 then what is the value of 2cos2a-1
Approach
We know that Sin2a+cos2a=1=> sin2a=1- cos2a….(eq1)
From the previous sum, we already know that
| LHS | RHS |
| sin4a-cos4a= | (sin2a-cos2a) |
| (1-cos2a-cos2a) ; applying eq.1 | |
| -2/3 (given in the Qs.) | =1-2cos2a |
| 2/3 | 2cos2a-1 ; multiplying both sides with (-1) |
Therefore, 2cos2a-1=2/3
Q. what is the value of (cosecA-sinA)2+(secA-cosA)2-(cotA-tanA)2
I’m dividing this equation into three parts. After that, I’ll apply the formula (a-b)2=a2-2ab+b2 on each of them
| (coseca-sina)2 | +(seca-cosa)2 | -(cota-tana)2 |
| Cosec2a-2cosecA x sinA+sin2a | +sec2a-2secA x cosA+cos2a | -(cot2A-2cotA x tanA +tan2A) |
If you observe the bold parts: 2 cosec A x sin A= 2 x 1 (because cosec and sin are inverse of each other so sin x cosec =1). Same situation will repeat two other parts (because sec x cos = 1 and tan x cot =1.)
So now our three parts will look like:
| Cosec2a-2+sin2a | +sec2a-2+cos2a | -(cot2A-2 +tan2A) |
=Cosec2a-2+sin2a+ sec2a-2+cos2a- cot2A+2 -tan2A ;donot make silly mistakes while opening brackets with minus sign e.g. -(a+b-c)=-a-b+c.
Now let’s club them according to the three trig. Formulas (e.g.sin2a+cos2a=1)
= (Cosec2a- cot2A) +(sin2a+ cos2a)+( sec2a -tan2A)+(-2-2+2)
=1+1+1-2 ;because each of those pair is equal to 1.
=1
Q. If sinA-cosA=7/13, what is the value of sinA+cosA, given that A is an acute angle?
we know that
(a-b)2=a2-2ab+b2
| LHS | RHS |
| (sinA-cosA)2= | Sin2a-2sinA x cosA +cos2A |
| (7/13)2= | 1-2sinA x cosA; because Sin2A+cos2A=1 |
Therefore, 2sinA x cosA=1-(7/13)2=….eq1
Moving forward, we also know that
(a+b)2=a2+2ab+b2
| LHS | RHS |
| (sinA+cosA)2= | Sin2a+2sinA x cosA +cos2A |
| 1+2sinA x cosA; because Sin2A+cos2A=1 | |
| 1+1-(7/13)2 ; plugging the value from eq.1 | |
| (sinA+cosA)2= | =289/169=(17/13)2 |
Taking square-root on both sides,
sinA+cosA=17/13, that’s our final answer.
Mock Questions
- Find value of Cos2a+(1/1+cot2a)
- 0
- 1
- 2
- 3
- Find value of 2sin2a+4sec2a+5cot2a+2cos2a-4tan2a-5cosec2a
- 0
- 1
- 2
- 3
- Find the value of (cosA-sinA)2+(cosA+sinA)2
- 0
- 1
- 2
- 3
- Find the value of Cot2A x (sec2A-1)
- 0
- 1
- 2
- 3
- Find the value of (secA x cotA)2 – (cosec A x cosA)2
- 0
- 1
- 2
- 3
- secA/(cotA+tanA) will be equal to
- cosA
- cosecA
- sinA
- tanA
- (1+tanA+secA)(1+cotA-cosecA) will be equal to
- 0
- 1
- 2
- 3
- Cos6A+sin6A=
- 1-3(cosA x sinA)2
- 1+3(cosA x sinA)2
- 1-3(cosA x sinA)3
- 1-3(cosA x sinA)6
- (sin2A x cos2B) – (cos2A x sin2B) will be equal to
- Sin2A-cos2A
- Sin2A+cos2A
- Cos2A – cos2A
- Sin2A-Sin2B
- (tanA+cotA)(secA-cosA)(cosecA-sinA)=
- 0
- 1
- 2
- 3
- If Tan2A+Tan4A=1 then what is the value of cos2A+cos4A, given that A is an acute angle?
- 0
- 1
- 2
- 3
- (cosecA-cotA)2=
- (1-cosA)/(2+cosA)
- (1+cosA)/(1-cosA)
- (1-cosA)/(1+cosA)
- (1-cosA)x(1+cosA)
- [(tanA+secA)2-1]/[(tanA+secA)2+1], will be equal to
- tanA
- secA
- sinA
- cosA
- [(sin220+sin270)/(sec250-cos240)]+2cosec258-2(cot58xtan32)-(4tan13 x tan37 x tan45 x tan53 x tan77)
- 0
- -1
- 1
- None of above
- If cotA=root7, what is the value of (cosec2A-sec2A)/( cosec2A+sec2A)
- 3/2
- 3/4
- 4/3
- 2/3
- (sinA+cosecA)2+(cosA+secA)2 will be equal to
- Tan2A+cot2A+4
- Tan2A-cot2A+5
- Tan2A-cot2A
- Tan2A+cot2A+7
Answers
1)b,2)b, 3)c, 4)b, 5)b, 6)c , 7)c, 8)a, 9)d, 10)b, 11)b, 12)c,13)d, 14)b, 15)b,16)d
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Good job…kudos….
Thanks Mrunalsir
Is there any twitter account of upsc chairman or its members ?
thanks sir…
Sir yesterday SSC released corrigendum regarding notification of SI in paramilitary forces
and IO in narcotics bureau.. it is mentioned that post of IO in NCB is cancelled..why is it so?
and is there any chances of separate notification for IO in NCB?
Is the post of NCB in SSC CGL same as that of this notification?
Mrunal Sir, u r making things so easy which would make the competition tougher as most of the candidates will be equally competent :D :D
in 9 ques: answer could be : cos^2(B) – cos^2(A), isnt it?
13 ques: i m getting answer (c) option- sinA
14 ques: there r 4 parts. i cud simplify 1,3,4 parts of it. but dont know what to do with 2nd part- 2cosec^2(58)
anyone solved the above ques, kindly comment. thank u so much.
Hahaha
Haha , same thoughts cross my mind sometimes but actually only Level 3 players (as per Mrunal) will able to retain everything.
Let the best take it all………..
tan(32) = cot(90-32) = cot 58.
that makes it cot^2(58)
now cosec^A = 1 + cot^A (take 2 common)
thanks HK. I got it where i was committing the mistake :). what abt 9 and 13ques?
if one is clever u dont need to solve 9th.. the 1st 3 options have 1 angle only. its quite unlikely that one angle gets easily eliminated. if want to solve. try this :
4th option has sin of A and B so just substitute (cosA)^2 = 1 – ((sinA)^2) and (cosB)^2 = 1 – ((sinB)^2).. next open up brackets that will come to option 4..
13th ans is SinA
(tan^2A + sec^2A + 2tanAsecA – 1)/(tan^2A + sec^2A + 2tanAsecA + 1)
put sec^2A in numerator = 1 + tan^2A
put tan^2A + 1 in denominator = sec^2A
(2tan^2A + 2tanAsecA)/(2sec^2A + 2tanAsecA)
take common and cancel we get tanA/secA = SinA
In trigonometrey, most of the questions we can solve very fast, by substitution of an angle…
for example, just observe the options and take an angle which gives different answers to different choices. In question 13, 30 degree angle is such an angle, and answer is SinA
HK, If you got some time then try to post Question 14 expression-1 , denominator, how that evaluate to = 1.
What is about the first part of the question number 14?
{ sec }^{ 2 }50-{ cos }^{ 2 }40={cosec} ^ { 2} 40-{cos}^{2}40
then what? I cannot solved the problem hence forth.
I mean I find the answer None of These but here in the answer section it is given -1 . Please consider.
if that term become cot^2(40) in the denominator of the first expression only then it is possible to achieve the -1 as the answer.Please clarify.
Question 14 ) To me, It seems some typing error in this question :-
as ,
last expression = – 4 , while ; 3rd expression = -2
first expression = (sin^2 20 + sin^2 70 )/(sec^2 50-cos^2 40) ;
thus here Numerator is 1;
so total first expression’s value seems < 1. AND
COSEC^2 58 = 2.77 approx , so summation of expression 1 and expression 2 comes out to be < 2.77 while we have "-6" from expressions- 3 and exp-4 , so either asnwer is none of these or there is typing error.
Although in question 14 ) expression-1 = -3 , epression-2+ expression-3 = 2 and expression-4 = -4 , which will give answer “-1”. but that is the same fallacy as ankit earlier asked.
In q. 9 .. ans is sin^a-sin^b..
sir i would request you to shift your focus back to Civil service exam … from many days i have seen many articles for SSC and other stuff but not a single article for CSE …
Hahaha….i can understand your frustation as i am from Maths background too…But these are also an integral part of CSE-Paper 2…and believe me these are very tough for those from non-maths background….
i dont think upsc focus much on maths in CSE paper 2…these questions are typical SSC questions…are u appearing this year??
Yup I am…
best of luck …my only advice is make sure that u r ready for the surprises…
I had filled the SSC CGLE 2013 form twice completely (fee, photo, sign everything). In the notification, it was specified that if a candidate fills multiple forms online, the latest complete form will be accepted. But now, sscwr.net has published the list of candidates with regn ids. In that list, they have accepted my first form.
Do anyone have any idea what to do now?
Had anybody else also filled form twice?
Even I submitted multiple entries but I didn’t face any problem. It could be some technical error or some mistake on your part but all I can say now is that nothing can be done to rectify it as it’s too late. My advice to you is-don’t lose hope,several other exams are coming out,you can prepare for them, if not, then try your luck with SSC CGL next year.
Good Luck!
Also,it was specified that you can send multiple entries in case of Part 1,the moment you submit part 2,it is deemed to be the final application and is accepted. Other option is-go to the site where you can download your admit card,search for your name by providing your d.o.b and see if there are multiple admit cards available in your name or not.
Did you complete part1 and part2 regn in all the multiple entries?
I filled multiple entries of part 1 only as there was some discrepancy but I only filled part 2 once and that was my final application.
IN CASE OF MULTIPLE APPLICATIONS FOR ONLINE APPLICATION, THE LAST APPLICATION FOR WHICH PART-I AND PART-II
REGISTRATION HAVE BEEN COMPLETED WILL BE ACCEPTED
Hey can someone help me with questions 6,7,9-16?? I feel like such a noob,I can’t solve them even after hours of hard work.
Please help!!
hey priyanka..
take it easy………these r vry simple…
q.6 just convert it into sin & cos and take LCM denom. it will be sinA
q.7 convert into sin & cos and multipjy it.
1 +cosA/SinA – 1/sinA + SinA/CosA +1 – 1/cosA+ 1/cosA + 1/sinA- 1/SinA.cosA
2+ cosA/SinA + SinA/CosA – 1/SinA.CosA
take LCM and u get answer = 2
q.9 sin^2A(1-sin^2B) – cos^2A.sin^2B
Sin^2A – Sin^2A.sin^2B – cos^2A.sin^2B
sin^2A – sin^2B(sin^2A + Cos^2A)
sin^2A – Sin^2B
q.16
sin^2A + Cosec^2A + 2.sinA.CosecA + cos^2A + sec^2A + 2.cosA.secA
(sin^2A + cos^2A) + (cosec^2A – 1) + (sec^2A – 1) +4 +2
7+ cot^2A + tan^2A ………………….
Thank you!
Dont try to prove it(LHS=RHS, Hence proved), because this is not some board exam. We can solve most of these problems by just plugging in the values for 30,45 or 60 degrees
More difficult the way u r suggesting, more calculations more chances of mistakes…. Plus even if u do every thing right answer may still be wrong as tan 45 equals cot45 and so and so
Dear priyana just convert everything into sin and cos and simple algebra
7th cos plus sin plus 1 by cos multiply by sin plus cos minus 1 by sin ie cos plus sin whole square minus 1 divided by sin multiply cos that become sin square plus cos square plus 2 sin cos minus 1 by sin cos… Now sin square plus cos square 1 so 2 sin cos by sin cos equals 2
I would recommend Vindos’s method..thanks Vinod for pointing out..
we’re not writing any board exam here..
take any angle say 45 degree
qn 7 becomes (1+1+root 2)(1+1-root2)=(2+root2)(2-root2)=4-2=2
simple…why to go thru all those equations..
In 9 th conver cos square into 1 minus sin square and simple algebra
Dil tod Diya tumne
11. take tan^2 common so it become tan^2 (1+tan^2) ie tan^2 *sec^2 ie sin^2/cos^4 so cos^4 =sin^2 substitute this it becomes sin^2 =cos^2 ie 1
u could still change your mind
ans. are
6. 1/COSA/COSA/SINA+SINA/COSA
=1/COSA/COS2A+SIN2A/SINA*COSA
=1/COSA*SINA/1*COSA
=SINA
7.(1+SINA/COSA+1/COSA)(1+COSA/SINA-1/SINA)
=[(COSA+SINA)+1/COSA][(SINA+COSA)-1/SINA]
=(SINA+COSA)2-1/COSA*SINA
=COS2A+SIN2A+2COSA*SINA-1/COSA*SINA
=1+2COSA*SINA-1/COSA*SINA
=2
9.SIN2A(1-SIN2B)-(1-SIN2A)SIN2B
=SIN2A-SIN2A*SIN2B-SIN2B+SIN2A*SIN2B
=SIN2A-SIN2B
16. SIN2A+COSEC2A+2*SINA*COSECA+COS2A+SEC2A+2*COSA*SECA
=SIN2A+COS2A+2*SINA*COSECA+2*COSA*SECA+COSEC2A+SEC2A
=1+2+2+1+COT2A+1+TAN2A
=7+COT2A+TAN2A
WHERE READ 2 AS SQUARE
Thanks a lot Dhanjit! :)
It would be great if you could also help me with questtions 10,11,12,13,14 & 15.
Hi,
Ques 10) (tan A+cot A)(sec A- cos A)(cosec A- sin A)
=(sin A/cos A +cos A/sin A)(1/cos A- cos A) ( 1/sin A- sin A)
= ( sin Sq A + cos Sq A)/(sin A + cos A) (1 -cos Sq A)/cos A (1 – sinSq A) / sin A.
Take sin Sq A * cos Sq A common then,
= ( sin Sq A Cos Sq A) /( Sin Sq A . Cos Sq A)
= 1. Option – B
Question 11) let t = tan Sq A , s = sin Sq A , c = cos Sq A.( read by this convention it makes it easy to type).
now question,
t^2 + t^4 = 1 ( given)
s^2/c^2 + s^4+c^4 = 1
take LCM ,
(s^2 * c^4 + c^2 * s^4)/ c^2 * c^4 = 1 ,
Take s^2 * c^2 , common in numerator ,
s^2 * c^2 ( c^2 + s^2)/c^2 * c^4 = 1
put s^2 * c^2 = 1 , and cancel out c^2 then ,
s^2 /c^4 = 1,
s^2 = c^4,
Now put s^2 = 1 – c^2 , in above equation ( that is sin Sq A = 1- cos sq A; read c = cos A and s = sin A , as told earlier),
we get ,
1- c^2 = c^4 ,
thus 1 = c^2 * c^4 or c^2 * c^4 = 1 , thus option – B
how it is possible 1-C^2=C^4 => 1=C^2*C^4?
Ques 12 ) ( cosec A – cot A)^2
= (1/ sin A – cosA / sin A)^2
Take 1/ sin^2 A common then,
= 1/ sin^2 A ( 1 – cos A)^2
put sin^2 A=1- cos^2 A , in above eqn then ,
= 1/ (1- cos^2 A) * (1-cosA)^2
Now apply a^2 – b^2 = (a+b)(a-b) on (1- cos^2 A)in denominator, we get ,
= 1 / ( 1- cos A) (1+ cos A) * ( 1- cos A)^2
= ( 1- cos A)/( 1 + cos A).
so Option – C
Question 13) read t = tan^2(A) s = sec^2(A)
Now question ,
((t+s)^2 – 1)/ ((t+s)^2 + 1)
open squares, and use sec^2A = 1+tan^2 A
= (t^2 + s^2 + 2 t*s – s^2 +t^2)/(t^2 + s^2 + 2 t*s + s^2 -t^2)
=(2 t^2 + 2 ts) /(2 s^2 + 2 ts)
take common or divide,
= (t/s) (( t + s) / (t + s))
= t / s
= (sin / cos )* (1/cos)
= SIN A , Option C.
Question 15) Cot A = root 7 ,
thus , let perpendicular = k , base = root 7 *k then , Hyp = root 50 *k,
this will give cosec^2A = 50 and sec^2 A = 50/7. put values and answer will be 3/4. option- B.
Question 14 ) To me, It seems some typing error in this question :-
as ,
last expression = – 4 , while ; 3rd expression = -2
first expression = (sin^2 20 + sin^2 70 )/(sec^2 50-cos^2 40) ;
thus here Numerator is 1;
so total first expression’s value seems < 1. AND
COSEC^2 58 = 2.77 approx , so summation of expression 1 and expression 2 comes out to be < 2.77 while we have -6 from expressions- 3 and exp-4 , so either asnwer is none of these or there is typing error.
sorry Numerator is 1 thus expression 1 is < 1.
Sorry its not the case Question 14) Expression-1 = -3 , expression-2 + expression-3 = 2 and expression-4 = -4 , so answer = -1.
Thanks a lot Fanish! :)
10. (SINA/COSA+COSA/SINA)(1/COSA-COSA)(1/SINA-SINA)
=(SIN2A+COS2A/COSA*SINA)(1-COS2A/COSA)(1-SIN2A/SINA)
=1/COSA*SINA*SIN2A/COSA*COS2A/SINA
=1
11.TAN2A(1+TAN2A)=1
=SIN2A/COS2A(1+SIN2A/COS2A)=1
=SIN2A/COS2A(COS2A+SIN2A/COS2A)=1
=SIN2A=COS4A
COS2A+COS4A=1-SIN2A+SIN2A=1
12.(1/SINA-COSA/SINA)2
=(1-COSA/SINA)2
=(1-COSA)(1-COSA)/(1+COSA)(1-COSA)
=(1-COSA)/(1+COSA)
13.[(TAN2A+SECA)2-(SEC2A-TAN2A)]/[(TAN2A+SEC2A+2SECA*TANA)+SEC2A-TAN2A]
=2TAN2A+2SECA*TANA/2SEC2A+2SECA*TANA
=SIN2A/COS2A+1/COSA*SINA/COSA/1/COS2A+1/COSA*SINA/COSA
=SIN2A+SINA/1+SINA
=SINA(1+SINA)/1+SINA=SINA
14.HERE SM MISTAKE IS THERE COS2A SHOULD E RPLACED WITH COT2A
SIN2(90-20)+SIN270/COSEC240-COT240+2COSEC258-2COT58*COT58-4*TAN45
=1/1+2(COSEC258-COT258)-4*1
=1+2-4=-1
15.B=ROOT7,P=1,H=ROOT8
(H/P)2-(H/B)2/(H/P)2+(H/B)2
=(ROOT8/1)2-(ROOT8/ROOT7)2/(ROOT8/1)2+(ROOT8/ROOT7)2
=8-8/7/8+8/7=56-8/56+8=48/64=3/4
READ 2 AS SQUARE
Thank you :)
mrunal please publish rest of the ECONOMIC survey and shift ur focus more on civil services it’s a request
Your every article is very much impressive. Keep it up. Thank you Mrunal Sir……
Thanks
mrunal bhai
remaining articles on Economic survey pls!
thanx a lot……..
good job
hi mrunal i am ur new subscriber and i think what you do is very good.
thanks
mrunal sir….. like the 4 parts of Trignometry series which u have given to us.. in same way will u give some articles on GEOMETRY… bcoz geometry is also asked in SSC CGL and contains many questions ….
pliz give an article on geometry..thanks for your great jobs
13. Im getting answer (c) option- sinA
14.the question should be like this
[(sin220+sin270)/(sec250-cot240)]+2cosec258-2(cot58xtan32)-(4tan13 x tan37 x tan45 x tan53 x tan77)
instead of
[(sin220+sin270)/(sec250-cos240)]+2cosec258-2(cot58xtan32)-(4tan13 x tan37 x tan45 x tan53 x tan77)
I want to ask that all these previous post including this is sufficient for cgls trigonometry? pls tell me
i think it should be…………………………………..
good job sir ,but please give more article related to upcoming upsc2013
Mrunal ji,
plz blocked some users who are spoiling this portal by making unnecessary comments.
plz explain percentsge change graphics
GREETINGS SIR PLZ SUGGEST SOME BOOKS FOR DATA SUFFICIENCY AND SERIES PROBLEMS……ITS TAKING MINUTES MORE TO SOLVE THAN REQUIRED……
Please somebody help me solve Ques. no. five.
Please somebody help me solve ques. no. 5
simple secA.cotA=(1/cosA)(cosA/sinA)=cosecA and cosecA.cosA=(1/sinA)(cosA)=cotA
Eqn changes now as cosec^2(A)-cot^2(A)=1 (identity)
A question for the people appearing for SSC CGl-How many questions do you manage to do in 2 hours (I’m referring to actual last year papers) and how many do you manage to get correct??
when i attempted first of the 8 papers of last 2 yr , i gt 125 n when i attempted 8th one, gt 148 (after negative marking and in given time constraint on an omr sheet)
So you’re doing 150 questions on an average I guess?
arnd 45 each in reasoning, maths n english… 25-30 in gk
Hey Priyanka,do u have last 5 yrs papers of CGL? Then plz give me link of that.
Last 3 year paper would be enough.
Here is the link:
http://sscportal.in/community/cgl/previous-year-papers
thanks dear :)
for those who have cgl-2013 on 21st.. is there any way to get question papers of exam conducted on 14th??
wish you a very prosperous new year sir
please don’t mar the image of this blog by writing unnecessary comments in which you are trying to impress someone or trying to prove your intelligence to someone..
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