Thursday, 21 April 2016

[Q] Trigonometry Tricks - 2

Please read Part-1 before reading this article.

This article covers some concepts that are taught in classes 11th and 12th, and hence new for arts/commerce candidates.

Few things which the above figure represents :

  • There are 4 quadrants (shown with I, II, III and IV).
    • Quadrant I - 0° to 90°
    • Quadrant II - 90° to 180°
    • Quadrant III - 180° to 270°
    • Quadrant IV - 270° to 360°

  • In the first quadrant, all the trigonometric functions are positive. So the values of sin56, cos18, tan89, cot67, cosec33, etc. are positive. Note that I have taken the angles 56, 18, 89, 67, 33 and all of them are less than 90(hence belong to the first quadrant) 
  • In the second quadrant, only sin and cosec are positive, and rest are negative. Hence sin91, sin135, cosec120, sin116, etc. are positive while cos135, cot120, tan95, etc, are negative. 
  • In the third quadrant, only tan and cot are positive, and rest are negative. Hence tan200, tan198, cot255, etc. are positive while cos255, sin220, cosec265, etc, are negative. 
  • In the fourth quadrant, only cos and sec are positive, and rest are negative. Hence cos300, cos350, sec290, sec285, etc. are positive while tan359, cot355, sin340, etc, are negative. 
  • The mnemonic to remember which trigonometric function is positive in which quadrant is - All Students Take Calculus. "All" is the first word of the sentence and hence represents the first quadrant. All trigonometric functions are positive in the 1st quadrant. Second initial is "S" which represents sin (indicating sin/cosec are positive in 2nd quadrant). Third initial is "T" which represents tan (indicating tan/cot are positive in 3rd quadrant). Fourth initial is "C" which represents cos (indicating cos/sec are positive in 4th quadrant).
  • Gist: 
    • All trigonometric functions are positive in the 1st quadrant
    • sin/cosec are positive in 2nd quadrant (sin and cosec are reciprocal of each other and hence their signs are same)
    • tan/cot are positive in 3rd quadrant
    • cos/sec are positive in 4th quadrant

Converting trigonometric functions :

sin(90 - A) = cosA, and hence sin65 = sin(90 - 25) = cos25
sin(90 + A) = cosA, and hence sin135 = sin(90 + 45) = cos45
cos(90 - A) = sinA, and hence cos85 = cos(90 - 5) = sin5
cos(90 + A) = -sinA, and hence cos135 = cos(90 + 45) = -sin45 = -(1/√2)
where A is any acute angle
  • sin(90 + A) refers to a value in the 2nd quadrant because A is an acute angle and hence (90 + A) would cover angles from 90 to 180 degrees (depending upon the value of A). 90 to 180 degrees is the range of 2nd quadrant. Now we have seen that in the second quadrant sin is positive. Hence sin(90 + A) = +cosA.
  • cos(90 + A) = -sinA, because in the second quadrant cos is negative.
  • (90 - A) represents the 1st quadrant and in the 1st quadrant, all the trigonometric functions are positive, hence:
    • sin(90 - A) = +cosA and hence sin75 = sin(90 - 15) = cos15 (here A = 15) 
    • cos(90 - A) = +sinA 
    • tan(90 - A) = +cotA 
    • cot(90 - A) = +tanA 
    • sec(90 - A) = +cosecA 
    • cosec(90 - A) = +secA
  • Now instead of 90 degrees if we have 180 degrees, then the functions are not converted. E.g. 
    • sin(180 - A) = sinA, and hence sin135 = sin(180 - 45) = sin45 
    • cos(180 - A) = -cosA and hence cos165 = cos(180 - 15) = -cos15 
    • tan(180 - A) = -tanA 
    • cosec(180 - A) = cosecA 
    • sec(180 - A) = -secA
  • Note that in the above lines only sin and cosec are positive, because (180 - A) represents 2nd quadrant and in the second quadrant only sin and cosec are positive.
Caution: While converting, please keep in mind that we check the sign of the function which is about to get converted. So if you want to convert sinX into cosY, first check where does X lie (in which quadrant), and then check the sign of "sin" (not cos) in that quadrant. If sin is positive in that quadrant, write sinX = +cosY, else write sinX = -cosY.

  • sin is converted into cos
  • tan is converted into cot
  • sec is converted into cosec
You need only this much knowledge to solve SSC questions [You don't have to do PhD after all :)]
Now let us solve some CGL questions:

Q. 1)

We have to convert sin3A into cos.
3A is an acute angle and hence sin3A lies in the 1st quadrant.
sin3A = +cos(90 - 3A)  [sin is positive in 1st quadrant, hence we have written +cos(90 - 3A)]
Hence, cos(90 - 3A) = cos(A - 26)
90 - 3A = A - 26 [Equating cos]
4A = 116
or A = 29
Answer : (A)
Note: In this question we had to convert sin into cos, hence we checked the sign of sin.

Q. 2)

cos20 = cos(90 - 70) = +sin70 [cos20 lies in the 1st quadrant and cos is positive in the 1st quadrant, hence we have written +sin70]
Hence, sin5θ = sin70
5θ = 70 [Equating sin]
θ = 14
Answer : (D)
Note: In this question, don't write sin5θ = cos(90 - 5θ), because this formula is applicable only for acute angles and 5θ is not necessarily an acute angle

I hope this concept of quadrants and conversion is clear.

Moving on, I discussed the basic trick of Trigonometry (putting the value of theta) in Part-1. Since this trick is extremely important, in each article of trigonometry I will solve some questions by assuming the value of θ so that you imbibe that method well. For reference, I am attaching the values-

Let us take some more questions from CGL-

Q. 3)

In this question you cant take θ = 45, 90 because that will make 2cos^4θ - cos^2θ = 0, and we know that denominator can never be zero. So you are left with two values: 0, 30 or 60
Take θ = 0
sec0 = 1
sin0 = 0
cos0 = 1
Put the above values and you will get the value of the expression as 1.
Answer: (A)

Q. 4)

Put A = 30
The value of the expression = 1/2/(1 + √3/2) + 1/2(1 - √3/2) = (2 - √3) + (2 + √3) = 4
Now put A = 30 in all the 4 options
(A) 4              (B) 4/√3                  (C) 1                  (D) √3
Answer : (A)
Note : Don't put A = 45 in this question, because then options A and B will give the same output.

Q. 5)

Put θ = 45
a = 0, b = 0
Value of the expression = (0 + 4)(0 - 1)^2 = 4
Answer : (A)

Q. 6)

We had seen earlier that when we have to assume two angles, it is best to assume them as 30 and 60.
α = 30 and β = 60
Then a = √3 and b = 1/√3
Then sin2β = 3/4
Put a = √3 and b = 1/√3 in all the 4 options and check which one of the2m is giving 3/4 as the output
Answer : (C)

As always, in case you have any doubt/doubts in this article, please drop a comment. Print a PDF of this article if you want to.
Part - 3 to follow soon...

Keep reading :)

To buy the super-hit SSC Hack-Book, follow the below link-
Buy SSC-Hack Book


  1. Jahapana tussi great ho............thought u will slow down a bit after the postponement of ssc cgl 2016.
    Also ek request hai,please dont avoid anonymous users comments as i have seen u didnt reply to any anonymous users(1-2 exceptions i think).may be they are the ones who are really interested in this blog but to stay anonymous inorder to avoid peer pressure & all:-).Anyways great work bro.

    1. haha...will keep that in mind :)

    2. are wah !! Sir has started replying to anonymous users too :-)

    3. are wah !! Sir has started replying to anonymous users too :-)

  2. Pls post more hacks on geometry number system mensuration and share your ways of preparation hard work everything which will be a motivation sir to all.tkfs

  3. Pls post more hacks on geometry number system mensuration and share your ways of preparation hard work everything which will be a motivation sir to all.tkfs

  4. This comment has been removed by the author.

  5. Sir in q 3:
    if we put θ=o then numerator become zero and we get sec^2θ=sec^2o=1^2=1

    is it right sir???

  6. Sir, I request you to update a post on Syllogism. I struggle a lot solving the questions and end up getting negative marks. It's quite a confusing topic for me.

    Looking forward to your reply.

  7. Replies
    1. Open the article by clicking on its title. At the bottom of the article, you will find a green "Print PDF" button...

    2. Super blog fact after reading some geometry hacks I started work in using that I solved quite easily....big thanks to u

  8. sir
    provide any pateform where we can share our douts just like facebook page where we can upload our douts picture.
    and you r doin great work sir thank you

  9. Hy sir, my prepartn s going on well by my own..but due to xam being extended I m looking to take coaching just to get myself polished so as to secure AIR below 500 .So cud u please recommend D teachers in Delhi.I want those who can take my doubts..I have done wid concepts properly .N also, sud I take only eng N maths class oR COMPLETE SSC BATCH would a wise decision. I regularly score 44-45 in previous papers in training in 25 min

  10. I score in reasoning I was mistyped

  11. sir plz upload few basic facts about co-ordinate geometry....

  12. plz give some hacks for surds n indices...

  13. This is a question from CGL Tier - II (2015) of AVERAGE. Sir can you please suggest a faster/good/better way/approach to solve this question?

    Q. The Average of 30 students of a class is 14 years 4 months. After admission of 5 new students in the class the average becomes 13 years 9 months. The youngest one of the five new students id 9 years 11 month old. The average age of the remaining 4 new students is

    (A) 11 years 2 months
    (B) 13 years 6 months
    (C) 12 years 4 months
    (D) 10 years 4 months

    1. 30*14(1/3) + 9(11/12) + 4a = 35*13(3/4)
      Solve for a

      And note that 14(1/3), 9(11/12) and 13(3/4) are mixed fractions

    2. Thank you so much Sir. :)

  14. Awesome wrk sir.... Being an employee, I couldn't attend coaching classes. But ur blog proved to me more than that of coaching classes. Good wrk...keep it up sir. Looking forward to learn more of the tricks.