### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

### Trigonometric, Identities & Formulas

Trigonometric, Identities & Formulas
Relation between angle and sides of the triangle, values of trigonometric functions at standard angles, transformation and trigonometric identities

TRIGONOMETRIC IDENTITIES & FORMULAS FOR  9TH, 10TH STANDARD

Trigonometry:
It is the branch of mathematics dealing with the relations of the sides and angles of right triangles.
$sin\theta =\frac{Perpendicular}{Hypotenuse}=\frac{P}{H}$$Cos\theta =\frac{Base}{Hypotenuse}=\frac{B}{H}$$Tan\theta =\frac{Perpendicular}{Base}=\frac{P}{B}$$Cot\theta =\frac{Base}{Perpendicular}=\frac{B}{P}$$Sec\theta =\frac{Hypotenuse}{Base}=\frac{H}{B}$$Cosec\theta =\frac{Hypotenuse}{Perpendicular}=\frac{H}{P}$From these six formulas we also find that $sin\theta =\frac{1}{cosec\theta }\Rightarrow cosec\theta =\frac{1}{sin\theta }$$cos\theta =\frac{1}{sec\theta }\Rightarrow sec\theta =\frac{1}{cos\theta }$$tan\theta =\frac{1}{cot\theta }\Rightarrow cot\theta =\frac{1}{tan\theta }$$Tan\theta =\frac{Sin\theta }{Cos\theta }\; \; and\; \; \; Cot\theta =\frac{Cos\theta }{Sin\theta }$ Shortcut method to learn these formulas
First of all learn this line: "Pandit Badri Parsad Har Har Bole Sona Chandi Tole" and then write these words in a tabular form as shown below
 Sona Chandi Tole Pandit Badri Parsad Har Har Bole
Convert  the first letter of the each word as shown  below, and in the last row write the reciprocal of sinθ, cosθ, and tanθ
 sin θ cos θ tan θ P B P H H B cosec θ sec θ cot θ

VALUE OF THE TRIGONOMETRIC FUNCTIONS WITH STANDARD ANGLE
 0o 30o 45o 60o 90o Sin 0 1/2 $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ 1 Cos 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ 1/2 0 Tan 0 $\frac{1}{\sqrt{3}}$ 1 $\sqrt{3}$ $\infty$ Cot $\infty$ $\sqrt{3}$ 1 $\frac{1}{\sqrt{3}}$ 0 sec 1 $\frac{2}{\sqrt{3}}$ $\sqrt{2}$ 2 $\infty$ cosec $\infty$ 2 $\sqrt{2}$ $\frac{2}{\sqrt{3}}$ 1

Method of finding these values
 Sin0o Sin30o Sin45o Sin60o Sin90o 0 1 2 3 4 $\frac{0}{4}$ $\frac{1}{4}$ $\frac{2}{4}$ $\frac{3}{4}$ $\frac{4}{4}$ 0 $\frac{1}{4}$ $\frac{1}{2}$ $\frac{3}{4}$ 1 $\sqrt{0}$ $\sqrt{\frac{1}{4}}$ $\sqrt{\frac{1}{2}}$ $\sqrt{\frac{3}{4}}$ $\sqrt{1}$ 0 $\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ 1

Steps For finding the  values of sinθ
1) Write counting from 0 to 4

2) Divide all the numbers by 4 and simplify these numbers

3) Taking square root of all these numbers

4) The values we get are the values on the sin function at different standard angles

For values of other trigonometric ratios
write all these values for sinθ in  the reverse order(from right to left)

For values of  tanθ use the formula tanθ = sinθ/cosθ

For values the values of cotθ use cotθ = 1/tanθ

For  the values of secθ use secθ = 1/cosθ

For the values of cosecθ use cosecθ = 1/sinθ

TRANSFORMATION OF TRIGONOMETRIC FUNCTIONS
sin(90 - θ)   = cosθ,                        cos(90 - θ) =  sinθ,
tan(90 - θ)  = cotθ,                          cot(90 - θ) = tanθ,
sec(90 - θ)   = cosecθ,                  cosec(90 - θ) = secθ

TRIGONOMETRIC IDENTITIES

In triangle ABC, by Pythagoras theorem

AB2 = AC2 +  BC2   ................  (1)

Dividing on both side by AB2

$\frac{AB^{2}}{AB^{2}}=\frac{AC^{2}}{AB^{2}}+\frac{BC^{2}}{AB^{2}}$$1=\left ( \frac{AC}{AB} \right )^{2}+\left ( \frac{BC}{AB} \right )^{2}$$1=\left ( \frac{P}{H} \right )^{2}+\left ( \frac{B}{H} \right )^{2}$$1=sin^{2}\theta +cos^{2}\theta$Dividing equation(1) by AC2  $\frac{AB^{2}}{AC^{2}}=\frac{AC^{2}}{AC^{2}}+\frac{BC^{2}}{BC^{2}}$ $\left ( \frac{AB}{AC} \right )^{2}=1+\left ( \frac{BC}{AC} \right )^{2}$$\left ( \frac{H}{P} \right )^{2}=1+\left ( \frac{B}{P} \right )^{2}$$cosec^{2}\theta =1+cot^{2}\theta \Rightarrow cosec^{2}\theta-cot^{2}\theta=1$ Dividing on both side by BC2$\frac{AB^{2}}{BC^{2}}=\frac{AC^{2}}{BC^{2}}+\frac{BC^{2}}{BC^{2}}$$\left(\frac{AB}{BC} \right )^{2}=\left ( \frac{AC}{BC} \right )^{2}+1$$\left(\frac{H}{B} \right )^{2}=\left ( \frac{P}{B} \right )^{2}+1$$sec^{2}\theta =tan^{2}\theta +1\Rightarrow sec^{2}\theta -tan^{2}\theta =1$
So we have the following results
$Sin^{2}\theta +Cos^{2}\theta = 1,\; \; \; Sin^{2}\theta=1-Cos^{2}\theta,\; \; \; Cos^{2}\theta =1-Sin^{2}\theta$
$Sec^{2}\theta -Tan^{2}\theta = 1,\; \; \; Sec^{2}\theta=1+Tan^{2}\theta,\; \; \; Tan^{2}\theta =Sec^{2}\theta-1$
$Cosec^{2}\theta -Cot^{2}\theta = 1,\; \; Cosec^{2}\theta=1+Cot^{2}\theta,\; \; \; Cot^{2}\theta = Cosec^{2}\theta-1$
Geometrical Representation of Trigonometric Ratios with the 30o, 60o .

Let ΔABC is an equilateral triangle. Therefore its each angle is = 60o . Draw AD⊥ BC so that AD bisect the base as well as the vertex angle A.
Let each side of ΔABC is a. That is  AB = BC = CA = a and BD = a/2 and ㄥA= 30o
In   ΔABD by Pythagoras theorem$AD^{2}=\sqrt{AB^{2}-BD^{2}}$$AD^{2}=\sqrt{a^{2}-\left (\frac{a}{2} \right )^{2}}$$AD^{2}=\sqrt{\frac{4a^{2}-a^{2}}{4}}$$AD=\frac{\sqrt{3}a}{2}$
$Sin 60^{o}=\frac{P}{H}=\frac{\frac{\sqrt{3}}{2}a}{a}=\frac{\sqrt{3}}{2}$$Cos 60^{o}=\frac{B}{H}=\frac{a/2}{a}=\frac{1}{2}$$tan 60^{o}=\frac{P}{B}=\frac{\frac{\sqrt{3}}{2}a}{\frac{1}{2}a}=\sqrt{3}$$cot60^{o}=\frac{B}{P}=\frac{\frac{1}{2}a}{\frac{\sqrt{3}}{2}a}=\frac{1}{\sqrt{3}}$$Sec 60^{o}=\frac{H}{B}=\frac{a}{a/2}=2$$Cosec60^{o}=\frac{H}{P}=\frac{a}{\frac{\sqrt{3}}{2}a}=\frac{2}{\sqrt{3}}$
Similarly we can find the geometrical representation of all trigonometric functions at 30o
Geometrical Representation of Trigonometric Ratios with the 45o.
In order to find the geometrical representation of trigonometric functions at 45o , we should make an isosceles right angled triangle . And then find the values with the same method as shown above.
Here let each equal side of the triangle = a
Then by pythagoras theorem $AB=\sqrt{BC^{2}+AC^{2}}$$AB =\sqrt{a^{2}+a^{2}} =\sqrt{2a^{2}}=\sqrt{2}a$$Sin45^{o} =\frac{P}{H}=\frac{a}{\sqrt{2}a} =\frac{1}{\sqrt{2}}$$Cos 45^{o} =\frac{B}{H}=\frac{a}{\sqrt{2}a}=\frac{1}{\sqrt{2}}$$Tan45^{o} =\frac{P}{B}=\frac{a}{a}=1$$Cot 45^{o} =\frac{B}{P}=\frac{a}{a}=1$$Cosec45^{o} =\frac{H}{P}=\frac{\sqrt{2}a}{a}=\sqrt{2}$$Sec45^{o} =\frac{H}{B}=\frac{\sqrt{2}a}{a}=\sqrt{2}$
Angle of elevation:

Angle of Depression:

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TRIGONOMETRY-CBSE Mathematics

### Comments

1. Good and very useful.