SSC CGL Trigonometry Notes Handwritten PDF Download

SSC CGL Trigonometry Notes Handwritten PDF Download

SSC CGL Trigonometry Notes Handwritten PDF Download.Today we are sharing Handwritten Trigonometry Notes for SSC exams. This Notes are of good quality and very important for forthcoming ssc exams. This PDF has been compiled by Shri Anil Kundalwal and proper credits goes to him. You may download this PDF from the link provided below.

 

 

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Important Notes & Short Tricks on Trigonometric Identities

Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important Notes & Short Tricks on Trigonometric Identities.

Important Short Tricks on Trigonometric Identities

Pythagorean Identities

  • sin2 θ + cos2 θ = 1
  • tan2 θ + 1 = sec2 θ
  • cot2 θ + 1 = csc2 θ

Negative of a Function

  • sin (–x) = –sin x
  • cos (–x) = cos x
  • tan (–x) = –tan x
  • csc (–x) = –csc x
  • sec (–x) = sec x
  • cot (–x) = –cot x

If A + B = 90o, Then

  • Sin A = Cos B
  • Sin2A + Sin2B = Cos2A + Cos2B = 1
  • Tan A = Cot B
  • Sec A = Csc B

For example:    

If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?

Solution:            

Tan A = Cot B, Tan A*Tan B = 1

So, A +B = 90o

(x+y)+(x-y) = 90o, 2x = 90o , x = 45o

Tan (2x/3) = tan 30o = 1/√3

If A – B = 90o, (A › B) Then

  • Sin A = Cos B
  • Cos A = – Sin B
  • Tan A = – Cot B

If A ± B = 180o, then

  • Sin A = Sin B
  • Cos A = – Cos B

If A + B = 180o                   

Then, tan A = – tan B

If A – B = 180o                    

Then, tan A = tan B

For example:    

Find the Value of tan 80o + tan 100o ?

Solution:Since 80 + 100 = 180

Therefore, tan 80o + tan 100o = 1

If A + B + C = 180o, then

Tan A + Tan B +Tan C = Tan A * Tan B *Tan C

sin θ * sin 2θ * sin 4θ = ¼ sin 3θ

cos θ * cos 2θ * cos 4θ = ¼ cos 3θ

For Example:What is the value of cos 20o cos 40o cos 60o cos 80o?

Solution: We know cos θ * cos 2θ * cos 4θ = ¼ cos 3θ

Now, (cos 20o cos 40o cos 80o ) cos 60o

¼ (Cos 3*20) * cos 60o

¼ Cos2 60o = ¼ * (½)2 = 1/16

If             a sin θ + b cos θ = m     &    a cos θ – b sin θ = n

then a2 + b2 = m2 + n2

For Example:

If 4 sin θ + 3 cos θ = 2 , then find the value of  4 cos θ – 3 sin θ:

Solution:

Let 2 cos θ – 3 sin θ = x

By using formulae a2 + b2 = m2 + n2

42 + 32 = 22 + x2

16 + 9 = 4 + x2

X = √21

If

sin θ +  cos θ = p     &     csc θ –  sec θ = q

then P – (1/p) = 2/q

For Example:

If sin θ + cos θ = 2 , then find the value of  csc θ – sec θ:

Solution:

By using formulae:

P – (1/p) = 2/q

2-(1/2) = 3/2 = 2/q

Q = 4/3 or csc θ – sec θ = 4/3

If

a cot θ + b csc θ = m     &    a csc θ + b cot θ = n

then b2 – a2  = m2 – n2

If

cot θ + cos θ = x     &    cot θ – cos θ = y

then x2 – y2 = 4 √xy

If

tan θ + sin θ = x     &    tan θ – sin θ = y

then x2 – y2 = 4 √xy

If

y = a2 sin2x + b2 csc2x + c

y = a2 cos2x + b2 sec2x + c

y = a2 tan2x + b2 cot2x + c

then,

ymin = 2ab + c

ymax = not defined

For Example:                    

If y = 9 sin2 x + 16 csc2 x +4 then ymin is:

Solution:            

For, y min = 2* √9 * √16 + 4

= 2*3*4 + 20 = 24 + 4 = 28

If            

y = a sin x + b cos x + c

y = a tan x + b cot x + c

y = a sec x + b csc x + c

then,     ymin = + [√(a2+b2)] + c

ymax = – [√(a2+b2)] + c

For Example:                    

If y = 1/(12sin x + 5 cos x +20) then ymax is:

Solution:            

For, y max = 1/x min

= 1/- (√122 +52) +20 = 1/(-13+20) = 1/7

Sin2 θ, maxima value = 1, minima value = 0

Cos2 θ, maxima value = 1, minima value = 0

Here are some important questions of Trigonometric identities.

(1)Value of image005 is

(a)image002

(b)image004

(c)image003

(d)None of these

Ans.(a)

image005 is equal to

image006

 

(2)If image007is acute and image008 then image009 is equal to

(a)

(b)3

(c)  2

(d)  4

Ans.  (c)

If sum of the inversely proportional value is 2

i.e if . image013 thenimage014

image015

 

so image009=2

or we can put image007= 45°

(3)The simplified value ofimage019 is

(a)-1

(b)0

(c)sec2x

(d)1

Ans. (d)

The simplified value of

image019is obtained by  putting x=y=45°

image022

(4) Find the value of image024

(a)  1

(b)  -1

(c)  2

(d)  -2

Ans. (c)

put image017

image026

(5) If image033 then image028 is equal

(a)7/4

(b) 7/2

(c)5/2

(d)5/4

Ans. (d)

image012 as we know thatimage034

j

on solving we get secimage007= 5/4

Note:if x+y=a

and x-y=b

then x=(a+b)/2 and y=(x-y)/2

Important notes on Triangles and their Properties

Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important notes on Triangles and their Properties.

Triangles and their properties

Area of triangle

Important notes on Triangles and their Properties

  1. When base and corresponding height is known: ½ * base * height = ½*c*h
  1. When all sides are given (Heron’s formulae): {s (s – a) (s – b) (s – c)}1/2  where, s = (a+b+c)/2
  1. When two sides and corresponding angle is given.: ½ a*c*sinθ
  1. When all the median are given (Median is a line joining the vertex to the opposite side at midpoint): 4/3 * {s(s – m1) (s – m2) (s –m3)}1/2

Note: Where, s = (m1+m2+m3)/2, m1,m2,m3 are three medians of the Triangle.

  1. When all the heights are given: 1/Area of ∆ = 4 {G (G – 1/h1) (G – 1/h2) (G – 1/h3),

Note: Where, G = ½ (1/h1 + 1/h2 + 1/h3)

Sine formulae of triangle

a/SinA = b/sinB = c/SinC = 2R Where, R is Circumradius

Cosine Formulae of triangle

CosA = b2 +c2 – a2 / 2bc

CosB= a2 +c2 – b2 / 2ac

CosC = b2 +a2 – c2 / 2ba

An interesting result based of cosine formulae

If in a triangle CosA = b2 +c2 – a2 / 2bc, Whereas b & c are the smaller sides then

Case I, b2 +c2  is greater than a2 then angle A is acute.

Case II, b2 +c2  is smaller than a2 then angle A is obtuse.

Case III, b2 +c2  is equals to a2 then angle A is Right Angle.

For example,

Ques: Find the type of the triangle ABC whose 3 sides are of length 11, 3 & 9.

Solution: Sum of square of two smaller sides is 32 + 92 = 9 + 81 = 90

Square of largest side is 112 = 121.

Hence, this triangle is obtuse angled triangle.

TRIANGLES AND ITS CERVICES

Medians of a Triangle

2

The medians of a triangle are line segments joining each vertex to the midpoint of the opposite side. The medians always intersect in a single point, called the centroid.

In the adjoining figure D, E, & F are midpoints of the side of the triangle while Line Segment AD, BE & CF are median. They are meeting at a common point I which is the centroid.

Properties:

  1. Centroid divides the Median in the ratio 2:1. i.e AI/ID = 2/1
  2. Apollonius Theoram:- To find the length of median when all sides are given: 4 * AD2 = 2(AC2 + AB2) – BC2
  1. All 3 median of a triangle divides the triangle in 6 equal parts.

i.e. ar of ∆AFI = ar of ∆AEI = ar of ∆BFI = ar of ∆BDI= ar of ∆CDI= ar of ∆CEI= ar of ∆ABC/6

Angle Bisectors

Important notes on Triangles and their Properties

Angle bisectors are the line segment which bisects the internal angles of the triangle. All the three angle bisectors meet at a common point called as Incentre.

In the adjoining figure Line Segment A1T1, A2T2, & A3T3 are Angle Bisector. While point I is Incentre of the triangle.

Incentre is the only point from which we can draw a circle inside the triangle which will touch all the sides of the triangle at exactly one point & this circle has a special name known as Incircle. And the radius of this circle is known as Inradius.

Properties:

  1. Inradius = Area of ∆ A1A2A3 / Semi perimeter of ∆ A1A2A3
  2. Angle formed at the incentre by any two angle bisector: Angle A1IA3  = Angle A1A2A3 /2) + 90o

Perpendicular Bisectors of Sides of the Triangle

Important notes on Triangles and their Properties
When the perpendicular bisectors of the side of the triangle is drawn they meet at a common point known as circumcentre.

This circumcentre is a special point as from this point we can draw a circle which will enclose the triangle in a way that all the vertex of the triangle lie on the circle. The radius of this circle is known as circumradius.

Properties:

In the adjoining figure DP, EP, & FP are perpendicular bisector of sides of the triangle. Point P is circumcentre.

  1. Circumradius = length of side AC * CB * BA / 4 * Area of triangle
  2. Angle formed at the circumcentre by any two linesegment joining circumcentre to the vertex: Angle APC = 2 * Angle ABC

Altitude of the Triangle

Important notes on Triangles and their Properties

The orthocenter of a triangle is the point where the three altitudes meet. This point may be inside, outside, or on the triangle.

In the adjoining figure AD, BE, & CF are three altitudes of a triangle. And point O is the orthocenter.

Properties:

  1. Angle BOC + Angle BAC = 180o

 

SIMILAR TRIANGLE

Important notes on Triangles and their Properties

 

If two triangles are similar, then the corresponding sides we have, shows the following relation:

  • The ratio of sides of triangle is proportional to each other.

Like AB/DE = BC/EF = CA/FD

For example:

If AB = 6cm, BC = 10cm & DE = 2cm. Find EF

Solution: AB/DE = BC/EF, 6/2 = 10/EF, Therefore EF = 2*10/6 = 10/3

  • The height, angle bisector, inradius & circumradius are proportional to the sides of triangle.

Median ∆ABC / Median ∆DEF = Height ∆ABC/ Height ∆DEF = AB/DE

  • The areas of the triangles are proportional to the square of the sides of the corresponding triangle.

Area ∆ABC/ Area ∆DEF = (AB/DE) 2

For example:

If AB = 6cm, DE = 10cm & Area of ∆ABC is 135 sqcm. Find Area of ∆DEF.

Solution: Area ∆ABC/ Area ∆DEF = (AB/DE) 2

135/ Area ∆DEF = (6/10)2 , Area ∆DEF = 135*(5/3)2 = 375 sqcm.

In a right angled triangle, the triangles on each sides of the altitude drawn from the vertex of the right angle to the hypotenuse are similar to each other and to the parent triangle.

Important notes on Triangles and their Properties

  • ∆ ABC ≈ ∆ ABD
  • ∆ ABC ≈ ∆ CBD
  • ∆ DBC ≈ ∆ ABD

Proof: In ∆ ABC & ∆ ABD. Angle B and Angle D is 90o. & Angle A is common in both. Hence by AA rule Both triangles are similar.

Some Important Question are as follows::

(1)In the given figure image001=?

image002

  1. 360°
  2. 720°
  3. 180°
  4. 300°

Ans.  (a)

Here in triangle AEC

image004

In triangle BFD

image006

Adding (i) and (ii) we get,

image001=360°

(2)In the given figure image008

image002

  1. 900°
  2. 720°
  3. 180°
  4. 540°

Ans.  (c)

In star like figure we calculate the sum of angle by using formula (n-4) 180, when n is the number of point in the star.

image010

image011

 

(3) In the given figure below, if AD = CD = BC, and image012=96°,How much is image013

image014

(a)  320

(b)  840

(c)   640

(d)  can’t be determined

Ans.  (c)

Here image016

image017

image018

image019

and from triangle ACB

image020

image021

(4)In the trapezium ABCD shown below, AD || BC and AB = 6, BC = 7, CD = 8, AD = 17, if sides AB and CD are extended to meet at E, find the measure of image022

image023

(a) 1200

(b)  1000

(c)   800

(d)  900

Ans.  (d)

image024

image025

(5)In the given figure, ABCD is a rhombus and AR = AB = BP, then the value ofimage026 is

(a)  600

(b)  900

(c)   1200

(d)  750

Ans.  (b)

image028

Let image029

image030

Similarly,

image031

(  ABCD is a rhombus)

image033

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