# SSC CGL Trigonometry Notes Handwritten PDF Download

## SSC CGL Trigonometry Notes Handwritten PDF Download

**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.

**Click Here**to Download TRIGONOMETRY Handwritten Notes for SSC Exams.

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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**

- sin
^{2}θ + cos^{2}θ = 1 - tan
^{2}θ + 1 = sec^{2}θ - cot
^{2}θ + 1 = csc^{2}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 = 90 ^{o}, Then**

- Sin A = Cos B
- Sin
^{2}A + Sin^{2}B = Cos^{2}A + Cos^{2}B = 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 = 90^{o}

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

Tan (2x/3) = tan 30^{o} = 1/√3

**If A – B = 90 ^{o}, (A **

**› B) Then**

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

**If A ****± B = 180 ^{o}, then**

- Sin A = Sin B
- Cos A = – Cos B

**If A + B = 180 ^{o} **

Then, tan A = – tan B

**If A – B = 180 ^{o} **

Then, tan A = tan B

**For example: **

Find the Value of tan 80^{o} + tan 100^{o} ?

**Solution:**Since 80 + 100 = 180

Therefore, tan 80^{o} + tan 100^{o} = 1

**If A + B + C = 180 ^{o}, 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 20^{o} cos 40^{o} cos 60^{o} cos 80^{o}?

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

Now, (cos 20^{o} cos 40^{o} cos 80^{o} ) cos 60^{o}

¼ (Cos 3*20) * cos 60^{o}

¼ Cos^{2} 60^{o }= ¼ * (½)^{2} = 1/16

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

then a^{2} + b^{2} = m^{2} + n^{2}

**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 a^{2} + b^{2} = m^{2} + n^{2}

4^{2} + 3^{2} = 2^{2} + x^{2}

16 + 9 = 4 + x^{2}

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 b^{2} – a^{2} = m^{2} – n^{2}

**If**

cot θ + cos θ = x & cot θ – cos θ = y

then x^{2} – y^{2} = 4 √xy

**If**

tan θ + sin θ = x & tan θ – sin θ = y

then x^{2} – y^{2} = 4 √xy

**If**

y = a^{2} sin^{2}x + b^{2} csc^{2}x + c

y = a^{2} cos^{2}x + b^{2} sec^{2}x + c

y = a^{2} tan^{2}x + b^{2} cot^{2}x + c

**then**,

y_{min} = 2ab + c

y_{max} = not defined

**For Example: **

If y = 9 sin^{2} x + 16 csc^{2} x +4 then y_{min} 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**, y_{min} = + [√(a^{2}+b^{2})] + c

y_{max} = – [√(a^{2}+b^{2})] + c

**For Example: **

If y = 1/(12sin x + 5 cos x +20) then y_{max} is:

**Solution: **

For, y max = 1/x min

= 1/- (√12^{2} +5^{2}) +20 = 1/(-13+20) = 1/7

**Sin ^{2} **

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

**Cos ^{2} **

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

**Here are some important questions of Trigonometric identities.**

(d)None of these

**Ans.(a)**

(2)If is acute and then is equal to

(a)

(b)3

(c) 2

(d) 4

**Ans. (c)**

If sum of the inversely proportional value is 2

(a)-1

(b)0

(c)sec^{2}*x*

(d)1

**Ans. (d)**

The simplified value of

is obtained by putting x=y=45°

(a) 1

(b) -1

(c) 2

(d) -2

**Ans. (c)**

(a)7/4

(b) 7/2

(c)5/2

(d)5/4

**Ans. (d)**

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__

**When base and corresponding height is known:**½ * base * height = ½*c*h

**When all sides are given (Heron’s formulae):**{s (s – a) (s – b) (s – c)}^{1/2}where, s = (a+b+c)/2

**When two sides and corresponding angle is given.:**½ a*c*sinθ

**When all the median are given**(Median is a line joining the vertex to the opposite side at midpoint): 4/3 * {s(s – m_{1}) (s – m_{2}) (s –m_{3})}^{1/2}

Note: Where, s = (m_{1}+m_{2}+m_{3})/2, m_{1},m_{2},m_{3 }are three medians of the Triangle.

**When all the heights are given:**1/Area of ∆ = 4 {G (G – 1/h_{1}) (G – 1/h_{2}) (G – 1/h_{3}),

**Note:** Where, G = ½ (1/h_{1} + 1/h_{2} + 1/h_{3})

**Sine formulae of triangle**

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

**Cosine Formulae of triangle**

CosA = b^{2} +c^{2} – a^{2} / 2bc

CosB= a^{2} +c^{2} – b^{2} / 2ac

CosC = b^{2} +a^{2} – c^{2} / 2ba

*An interesting result based of cosine formulae*

If in a triangle CosA = b^{2} +c^{2} – a^{2} / 2bc, Whereas b & c are the smaller sides then

**Case I, **b^{2} +c^{2} is greater than a^{2} then angle A is acute.

**Case II,** b^{2} +c^{2} is smaller than a^{2} then angle A is obtuse.

**Case III,** b^{2} +c^{2} is equals to a^{2} 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 3^{2} + 9^{2} = 9 + 81 = 90

Square of largest side is 11^{2} = 121.

Hence, this triangle is obtuse angled triangle.

__TRIANGLES AND ITS CERVICES__

**Medians of a Triangle**

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:*

- Centroid divides the Median in the ratio 2:1. i.e AI/ID = 2/1
- Apollonius Theoram:- To find the length of median when all sides are given: 4 * AD
^{2 }= 2(AC^{2}+ AB^{2}) – BC^{2}

- 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**

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 A_{1}T_{1}, A_{2}T_{2}, & A_{3}T_{3} 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:*

**Inradius**= Area of ∆ A_{1}A_{2}A_{3}/ Semi perimeter of ∆ A_{1}A_{2}A_{3}**Angle formed at the incentre by any two angle bisector:**Angle A_{1}IA_{3}= Angle A_{1}A_{2}A_{3}/2) + 90^{o}

**Perpendicular Bisectors of Sides of the Triangle**

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.

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

**Altitude of the Triangle**

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:*

- Angle BOC + Angle BAC = 180
^{o}

__SIMILAR__ __TRIANGLE__

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.

- ∆ ABC ≈ ∆ ABD
- ∆ ABC ≈ ∆ CBD
- ∆ DBC ≈ ∆ ABD

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

**Some Important Question are as follows::**

- 360°
- 720°
- 180°
- 300°

**Ans. (a)**

Here in triangle AEC

In triangle BFD

Adding (i) and (ii) we get,

- 900°
- 720°
- 180°
- 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.

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

(a) 32^{0}

(b) 84^{0}

(c) 64^{0}

(d) can’t be determined

**Ans. (c)**

and from triangle ACB

**(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

(a) 120^{0}

(b) 100^{0}

(c) 80^{0}

(d) 90^{0}

**Ans. (d)**

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

(a) 60^{0}

(b) 90^{0}

(c) 120^{0}

(d) 75^{0}

**Ans. (b)**

Similarly,

( ABCD is a rhombus)

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