### Mathematics Assignments | PDF | 8 to 12

PDF Files of Mathematics Assignments From VIII Standard to XII Standard PDF of mathematics Assignments for the students from VIII standard to XII standard.These assignments are strictly according to the CBSE and DAV Board Final question Papers

### 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

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 sinÎ¸, cosÎ¸, and tanÎ¸
 sin Î¸ cos Î¸ tan Î¸ P B P H H B cosec Î¸ sec Î¸ cot Î¸

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 sinÎ¸
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 sinÎ¸ in  the reverse order(from right to left)

For values of  tanÎ¸ use the formula tanÎ¸ = sinÎ¸/cosÎ¸

For values the values of cotÎ¸ use cotÎ¸ = 1/tanÎ¸

For  the values of secÎ¸ use secÎ¸ = 1/cosÎ¸

For the values of cosecÎ¸ use cosecÎ¸ = 1/sinÎ¸

TRANSFORMATION OF TRIGONOMETRIC FUNCTIONS
sin(90 - Î¸)   = cosÎ¸,                        cos(90 - Î¸) =  sinÎ¸,
tan(90 - Î¸)  = cotÎ¸,                          cot(90 - Î¸) = tanÎ¸,
sec(90 - Î¸)   = cosecÎ¸,                  cosec(90 - Î¸) = secÎ¸

TRIGONOMETRIC IDENTITIES

## Derivation of Sin2Î¸ + cos2Î¸ = 1

Let us take an right angled triangle ABC as shown below

In triangle ABC, by Pythagoras theorem

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

Dividing on both side by AB2

$\frac{AB^{2}}{AB^{2}}=\frac{AC^{2}}{AB^{2}}+\frac{BC^{2}}{AB^{2}}$

$1=\left ( \frac{AC}{AB} \right )^{2}+\left ( \frac{BC}{AB} \right )^{2}$

$1=\left ( \frac{P}{H} \right )^{2}+\left ( \frac{B}{H} \right )^{2}$

$1=sin^{2}\theta +cos^{2}\theta$

Derivation of cosec2Î¸ - cot2Î¸ = 1

Dividing equation(1) by AC2

$\frac{AB^{2}}{AC^{2}}=\frac{AC^{2}}{AC^{2}}+\frac{BC^{2}}{BC^{2}}$

$\left ( \frac{AB}{AC} \right )^{2}=1+\left ( \frac{BC}{AC} \right )^{2}$

$\left ( \frac{H}{P} \right )^{2}=1+\left ( \frac{B}{P} \right )^{2}$

$cosec^{2}\theta =1+cot^{2}\theta \Rightarrow cosec^{2}\theta-cot^{2}\theta=1$

Derivation of Sec2Î¸ - tan2Î¸ = 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  60o .

Let Î”ABC 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 Î”ABC is a. That is  AB = BC = CA = a and BD = a/2 and ã„¥A= 30o

In   Î”ABD by Pythagoras theorem

$AD^{2}=\sqrt{AB^{2}-BD^{2}}$

$AD^{2}=\sqrt{a^{2}-\left (\frac{a}{2} \right )^{2}}$

$AD^{2}=\sqrt{\frac{4a^{2}-a^{2}}{4}}$

$AD=\frac{\sqrt{3}a}{2}$

Now In   Î”ABD

$Sin 60^{o}=\frac{P}{H}=\frac{\frac{\sqrt{3}}{2}a}{a}=\frac{\sqrt{3}}{2}$

$Cos 60^{o}=\frac{B}{H}=\frac{a/2}{a}=\frac{1}{2}$

$tan 60^{o}=\frac{P}{B}=\frac{\frac{\sqrt{3}}{2}a}{\frac{1}{2}a}=\sqrt{3}$

$cot60^{o}=\frac{B}{P}=\frac{\frac{1}{2}a}{\frac{\sqrt{3}}{2}a}=\frac{1}{\sqrt{3}}$

$Sec 60^{o}=\frac{H}{B}=\frac{a}{a/2}=2$

$Cosec60^{o}=\frac{H}{P}=\frac{a}{\frac{\sqrt{3}}{2}a}=\frac{2}{\sqrt{3}}$

## Geometrical Representation of Trigonometric Ratios with the  30o .

Now In   Î”ABD

$Sin 30^{o}=\frac{P}{H}=\frac{a/2}{a}=\frac{1}{2}$

$Cos 30^{o}=\frac{B}{H}=\frac{\left ( \sqrt{3}/2 \right )a }{a}=\frac{\sqrt{3}}{2}$

$tan 30^{o}=\frac{P}{B}=\frac{a/2 }{\left ( \sqrt{3}/2 \right )a}=\frac{1}{\sqrt{3}}$

$cot 30^{o}=\frac{B}{P}=\frac{\frac{\sqrt{3}}{2}a}{\frac{1}{2}a}=\sqrt{3}$

$sec30^{o}=\frac{H}{B}=\frac{a}{\frac{\sqrt{3}}{2}a}=\frac{2}{\sqrt{3}}$

$Cosec \: 30^{o}=\frac{H}{p}=\frac{a}{a/2}=2$

## 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.