Important questions on Trigonometry with tricks for ssc

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Trigonometry is yet another scoring section of CGL. Most of the questions can be solved with jugaad.

Just like Algebra, where we assumed the values of variables, in trigonometry we will assume the value of 'theta'. And like algebra, make sure the value you are assuming for theta will not make the denominator zero.
How to assume value -
1. When you don't have to deal with fractions you can assume ϴ = 90 or 0
E.g. (acosϴ - bsinϴ)
2. When fraction is given and putting ϴ = 90 or 0 is making denominator zero, then you can go with ϴ = 45.
3. Don't assume a value for theta at which the trigonometric function is not defined. E.g. When tanϴ is given, you can't assume ϴ = 90.
4. When you are assuming two angles, go with A = 60 and B = 30
Note : These are not hard and fast rules and you can assume any value you like, but make sure denominator ≠ 0. Sometimes when you assume ϴ, you may end up with two options(say A and B) that are giving similar results (but two options will still get eliminated, i.e., C and D). Now change the value of theta and check only A and B.


 You only need to memorize the values of sin, cos and tan (for ϴ = 0, 30, 45 and 60). The values of cosec, sec and cot can be obtained by reciprocating sin, cos and tan respectively.
Now let's solve CGL questions

Q . 1. 

Put ϴ = 90

x = a(1 + 0) = a

y = b(1 - 0) = b

So x²/a² + y²/b² = 2

Answer : C

Some complex questions can be solved by assuming the value of theta

Q . 2.

Again assume ϴ = 90
Then a*1 + b*0 = cor a = c
You have to find the value of acosϴ - bsinϴ
acosϴ - bsinϴ = -b (Since ϴ = 90)
Now put a = c in all the 4 options to check which one can give '-b' as the output

Answer : D

Q . 3.

Assume ϴ = 45

a (tan45 + cot45) = 1

So a = 1/2

sin45 + cos45 = b

So b = √2

Put a = 1/2 and b = √2 in all the options and check which among the 4 equations is right (i.e. LHS should be equal to RHS)

Answer : A (both LHS and RHS are equal to 1)

Q. 4.

Assume ϴ = 45
(tanA - secA - 1)/(tanA + secA + 1) = -√2/(2 + √2) = -1/(√2 + 1) = 1 - √2 (rationalize)
Put ϴ = 45 in all the 4 options and check which one will give (1 - √2) as the output
A) √2  - 1
B) √2 + 1
C) 1 - √2
D) √2 - 1
Answer : C

Sometimes you would need to assume two angles

Q . 5

Let A = 60 and B = 30
Then n = 3 and m = √3
cos²A = 1/4 (since A = 60)
Now put n = 3 and m = √3 in all the options and check which one will give 1/4
Answer : B


There is one type of question which is frequently asked by SSC - 

Q . 6.

When you see secA + tanA = something (let's say 'p')   ... (1)
you can write, secA - tanA = 1/p                           ... (2)
Now add (1) and (2)
2secA = p + 1/p
secA = (p² + 1)/2p [You can memorize this formula]
tanA = (p² - 1)/2p

In the above question, p=2
So secA = 5/4
Now we have to find sinA. The best way to determine the value of a trigonometric function when the value of other function is given, is by making a triangle.
secϴ = Hypotenuse/Base
Here secA = 5/4, hence hypotenuse = 5 and base = 4, which means perpendicular = 3
sinA = perpendicular/hypotenuse = 3/5 = 0.6
Answer : C




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
Explanation

• 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.

Gist:

• 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)

 

 

Sine and Cosine Rules

See the image above and mug the formulas thoroughly. Sin and cosine rules are important in trigonometry and can help you in solving some complex questions.

Q. 1)

The figure will look something like this -

In triangle ACD
sin∠CAD/3x = sin45/AD   ... (1)
In triangle ABD
sin∠BAD/x = sin60/AD    ...  (2)

Divide equation (2) by (1)

3* sin∠BAD/sin∠CAD = sin60/sin45

sin∠BAD/sin∠CAD = √6/6 = 1/√6

Answer: (C)

Similarly you can use cosine law to find any angle, if all the sides are given or to find a side, if the other two sides and an angle is given.

Q. 2)  In the below figure, ABC is right angled at B and AD = CD. If ∠ACB=30, find ∠ABD

          (A) 30                (B) 60                (C) 45                (D) 75

∠BAD = 180 - (90 + 30) = 60

In triangle ABD

sin∠ABD/AD = sin60/BD  [Since ∠BAD = 60]

In triangle BCD

sin∠CBD/CD = sin30/BD

Divide equation (2) by (1)

sin∠CBD/sin∠ABD = sin30/sin60     [AD = CD and hence they will cancel out]

Now, sin∠CBD = sin(90 - ∠ABD) = cos∠ABD

Hence, cos∠ABD/sin∠ABD = sin30/sin60

cot∠ABD = 1/√3

Hence ∠ABD = 60

Answer: (B)

Alternative Method

The figure given in this question is very important and at times it is embedded in some other figures. There is one short-cut  to calculate the angle.

Imagine the triangle in circumscribed in a circle

Now, D will be the centre of the triangle with diameter AC. We can say with surety that AC is the diameter of the circle because ∠ABC = 90, and we know angle in a semicircle is right angle. Moreover AD = CD, hence D is the midpoint of the diameter or the centre of the circle

Now you can see that AD, CD and BD are radii of the circle. Hence AD = CD = BD

∠BAD = 180 - (90 + 30) = 60

∠ABD = ∠BAD = 60 [Since AD = BD]

Answer: (B)

Now let us see a CGL question, in which the above figure was embedded.

Q. 3) G is the centroid of Triangle ABC, and AG=BC. Find angle BGC.

(A) 60              (B) 90               (C) 120                (D) 75

Let AG = 2x
Then BG = x (centroid divided the median in 2:1 ratio)
BC = AG = 2x
Let AG when extended cuts BC at D
Then D is the midpoint of BC  (as AD is the median)
BD = DC = x  [Since BC = 2x]
Now DG = BD = DC = x
That means D is the centre of a circle with diameter BC and one of the radius as DG.
Hence BGC = 90 (angle in a semi-circle)
Answer: (B)

Q. 4)

sec²x + tan²x = 5/3

We know, sec²x – tan²x = 1

Adding the above two equations

2sec²x = 8/3 or sec²x = 4/3

secx = 2/√3

That means x = 30

cos2x = cos60 = 1/2
Answer: (C)
Method 2:
sec²x = 4/3
That means, cos^2x = 3/4 and sin^2x = 1 - 3/4 = 1/4
cos2x = cos^2x - sin^2x = 3/4 - 1/4 = 1/2

Q. 5)

Multiply and Divide RHS by 2

(cosx - sinx)/(cosx + sinx) = (√3/2 - 1/2)/(√3/2 + 1/2)

Match RHS with LHS and you can easily see x = 30

Answer: (A)

Method 2

Cross multiply

(cosx - sinx)(√3 + 1) = (cosx + sinx)(√3 - 1)

Solve it and you will get, tanx = 1/√3

Hence x = 30

Q. 6)

Put θ = 45
2y*cos45 - x*sin45 = 0
2y = x  ... (1)
2x*sec45  - y*cosec45 = 3
2x - y = 3/√2
4y - y = 3/√2  [Put x = 2y]
y = 1/√2
Hence, x = 2y = √2
x^2 + 4y^2 = 2 + 4*(1/2) = 2 + 2 = 4
Answer: (C)


Some important values to mug:

1.  sin15 = (√3 - 1) /2√2
2.  cos15 = (√3 + 1) /2√2
3.  tan15 = 2 - √3
4.  cot15 = 2 + √3

These values are very important to solve some tricky trigonometric questions. Examples :

Q. 7) 

Although you can solve this question with the direct formula, which I discussed in Geometry Tricks - 1. But let us assume, you forget that formula. In such cases, the values of sin15 and cos15 will come handy.

Let the perpendicular and base of the triangle be P and B, respectively.

sin15 = Perpendicular/Hypotenuse = P/100

P = sin15*100
Similarly, B = cos15*100

Area = 1/2 * P * B = 1/2 * sin15 * 100 * cos15 * 100 =  (√3 - 1) /2√2 * (√3 + 1) /2√2 * 100 * 100/2

Area = 100*100/8 = 1250

Answer: (D)

Q. 8)

       A) 0               B) 1                 C) -1                 D) 2

Put x = 15
= cot15/(cot15 - cot45) + tan15/(tan15 - tan45)
= (2 + √3)/(1 + √3) + (2 - √3)/(1 - √3)
= 1
Answer: (B)

[Maximum and Minimum Values of Trigo-Functions]

Few points to remember:

1. sinθ and cosθ both have "1" as their maximum value and "-1" as their minimum value. Hence the values of sinx, sin2x, cosx, cos3x, etc. lie between -1 and 1.
2. For sin²x and cos²x
• Minimum value = 0
• Maximum value = 1
3. For sinxcosx
• Minimum value = -1/2
• Maximum value = 1/2
4. Minimum value of (sin θ cos θ)n = (-1/2)n

Some SSC CGL questions :

      Q. 1) What is the least value of 2sin²θ + 3cos²θ

       (A) 1            (B) 2             (C) 3            (D) 0

      Since 2 is less than 3
      Minimum value = 2

      Answer: (B)

       

     Hence the least value of 4sec²θ + 9cosec²θ = 13 + 12 = 25

     Answer: (C)

      Q. 4) The maximum of 3sinx - 4cosx is

       (A) -1               (B) 5                (C) 7               (D) 9

       Maximum value = √(3^2 + 4^2) = √25 = 5
       Answer: (B)

Sometimes they ask the minimum/maximum value of a function. In CGL, that function would always be quadratic (ax² + bx + c). Please note that a quadratic function can't have both maximum and minimum values. 

• If "a" is positive, then the quadratic function will only have a minimum value. The maximum value would be infinite.
• If "a" is negative, then the quadratic function will only have a maximum value. The minimum value would be infinite.

But the process to find both minimum and maximum values is same. Hence you shouldn't be worried about the words "maximum" or "minimum". When finding the minimum/maximum value of a function, we use the concept of "differentiation". Although differentiation is a wide topic in itself, but for CGL purpose, we only have to learn the basics.

Just remember following things:

1. Differentiation of axⁿ = a*n*x^n-1. Hence differentiation of 4x³ = 12x² and differentiation of 3x² = 6x
2. Differentiation of ax = a. Hence differentiation of 4x is 4.
3. Differentiation of any constant is zero. Hence differentiation of 5 is 0.
4. If you have to differentiate ax² + bx + c, just differentiate each of its term separately and add the result.
5. Process of finding the minimum value of a function :
1. Differentiate the function
2. What ever result you get, equate it with zero
3. Find the value of x
4. Put this value of x in the original function to get the minimum value.

Let us solve a CGL question:

Q. 5) Find the minimum value of (x - 2)(x - 9)
(A) -11/4             (B) 49/4             (C) 0            (D) -49/4

(x - 2)(x - 9) = x^2 - 11x + 18

Now we will differentiate (x^2 - 11x + 18)

Note that in this question, the value of a is positive, i.e., +1 and hence the question has asked you the "minimum" value. Had the value of "a" been negative (like -1), they would have asked you the "maximum" value.

Hence, differentiation of (x^2 - 11x + 18) = 2x - 11

Now 2x - 11 = 0

x = 11/2

Put x = 11/2 in (x^2 - 11x + 18) to get the minimum value

Minimum value of x^2 - 11x + 18 = (11/2)^2 - 11*(11/2) + 18 = -49/4

Answer: (D)

 

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