- Trignometry concepts
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- Trigonometric Ratio’s
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- Trigonometric Ratio’s of Common angles
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- Trigonometric ratio’s of complimentary angles
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- Trigonometric identities

- Trigomometry from Greek
*trigõnon*, "triangle" and metron, "measure") is a branch of mathematics that studies relationships involving lengths and angles of triangles. - The field emerged during the 3
^{rd}century BC from applications of geometry to astronomical studies. - Trigonometry is most simply associated with planar right angle triangles (each of which is a two-dimensional triangle with one angle equal to 90 degrees).
- The applicability to non-right-angle triangles exists, but, since any non-right-angle triangle (on a flat plane) can be bisected to create two right-angle triangles, most problems can be reduced to calculations on right-angle triangles. Thus the majority of applications relate to right-angle triangles

- In a right angle triangle ABC where B=90°

- We can define following term for angle A

- We can define the trigonometric ratios for angle A as

*Notice that each ratio in the right-hand column is the inverse, or the reciprocal, of the ratio in the left-hand column.*

The reciprocal of sin A is cosec A ; and vice-versa.

The reciprocal of cos A is sec A

And the reciprocal of tan A is cot A

**These are valid for acute angles.** - We can define $tan A = \frac {sin A}{cos A}$

And $Cot A =\frac {cos A}{ Sin A}$

**Important Note**

Since the hypotenuse is the longest side in a right triangle, the value of

sin A or cos A is always less than 1 (or, in particular, equal to 1).

- Similarly we can have define these for angle $C$

We can define the trigonometric ratios for angle $C$ as

$sin C= \frac {Perpendicular}{Hypotenuse}=\frac {AB}{AC}$

$cosec C= \frac {Hypotenuse}{Perpendicular} =\frac {AC}{AB}$

$cos C= \frac {Base}{Hypotenuse}=\frac {BC}{AC}$

$sec C= \frac {Hypotenuse}{Base}=\frac {AC}{BC}$

$tan C= \frac {Perpendicular}{Base}=\frac {AB}{BC}$

$cot C= \frac {Base}{Perpendicular}=\frac {BC}{AB}$

We can find the values of trigonometric ratio’s various angle

- From the table we can see that cosec 0 is undefined. It is undefined as cosec A= hyp/Perp,when angle becomes zero, Perpendicular becomes zero and x/0 is undefined
- Similarly for Cot 0
- From the table we can see that ,tan 90 is undefined . It is undefined as tan A=perp/base ,when angle becomes 90,base becomes zero and x/0 is undefined
- Similarly , it can be explained for sec 90

State whether the following are true or false. Justify your answer.

(i) The value of $tan A$ is always less than 1.

(ii) $sec A =\frac {12}{5}$ for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) tan A is the product of tan and A.

(v) $sin A =\frac {5}{3}$ for some angle

i) False . The value of tan A increase from 0 to infinity

ii) True . The value of sec A increase from 1 to infinity

iii) False .Cos A is the abbreviation used for the cosine of angle A

iv) False .cot A is one symbol. We cannot separate it

v) False. The value of sin AA always lies between 0 and 1 and 5/3 > 1

The value of (sin30 + cos30) - (sin60 + cos60) is

(A) - 1

(B) 0

(C) 1

(D) 2

Answer (B)

In a right angle triangle ABC

$A+B+C=180$

Now $B=90$

So $A +C =90$

Or $C=90-A$

We have seen in Previous section the value for trigonometric ratios for angle $C$

$sin C= \frac {Perpendicular}{Hypotenuse}=\frac {AB}{AC}$

$cosec C= \frac {Hypotenuse}{Perpendicular} =\frac {AC}{AB}$

$cos C= \frac {Base}{Hypotenuse}=\frac {BC}{AC}$

$sec C= \frac {Hypotenuse}{Base}=\frac {AC}{BC}$

$tan C= \frac {Perpendicular}{Base}=\frac {AB}{BC}$

$cot C= \frac {Base}{Perpendicular}=\frac {BC}{AB}$

This can be rewritten as

$sin (90-A) =\frac {AB}{AC}$

$cosec (90-A) =\frac {AC}{AB}$

$cos (90-A) =\frac {BC}{AC}$

$sec (90-A)=\frac {AC}{BC}$

$tan (90-A)=\frac {AB}{BC}$

$cot( 90-A) =\frac {BC}{AB}$

Also we know that

$sin A= \frac {BC}{AC}$

$cosec A= \frac {AC}{BC}$

$cos A= \frac {AB}{AC}$

$sec A = \frac {AC}{AB}$

$tan A= \frac {BC}{AB}$

$cot A =\frac {AB}{BC}$

$Sin (90-A) =cos(A)$

$Cos(90-A) = sin A$

$tan(90-A) =cot A$

$sec(90-A)= cosec A$

$cosec (90-A) =sec A$

$cot(90- A) =tan A$

If a is the hypotenuse and b and c are other two sides,then

$a^2= b^2+ c^2$

This same theorem can be used to prove the below trigonometric identities

$Sin^2 A + cos^2 A =1$

$1 + tan^2 A =sec^2 A$

$1 + cot^2 A =cosec^2 A$

- Learn well the formulas for Trigonometric identities, trigonometric ratios, reciprocals The better you know the basic identities, the easier it will be to recognize what is going on in the problems.
- Generally RHS( Right hand side) would be more complex. So start from there and simplify it to the same form as LHS(Left hand side)
- It becomes many times easy, if Convert all sec, csc, cot, and tan to sin and cos. Most of this can be done using the quotient and reciprocal identities.
- If you see power 2 or more, it will involve using the below identities mostly

$Sin^2 A + cos^2 A =1$

$1 + tan^2 A =sec^2 A$

$1 + cot^2 A =cosec^2 A$ - Once you get the hang of it, you will begin to see patterns and it will be easy to solve these Trigonometric identities Problems
- Practice and Practice. You will soon start figuring out the equation and there symmetry to resolve them fast

Prove that $(sin^4 \theta - cos^4 \theta +1) cosec^2 \theta = 2$

L.H.S. = $(sin^4 \theta - cos^4 \theta +1) cosec^2 \theta$

$= [(sin^2 \theta - cos^2 \theta) (sin^2 \theta + cos^2 \theta) + 1] cosec^2 \theta$

$= (sin^2 \theta - cos^2 \theta) + 1) cosec^2 \theta$

[Because $Sin^2 A + cos^2 A =1$]

$= 2sin^2 \theta cosec^2 \theta$ [Because $1- cos^2 \theta = sin^2 \theta$ ]

= 2 = RHS

3. Reciprocal of the trigonometric ratio cosecant

5. Value of cosine of 0 degrees

6. Ratio obtained by dividing sine by cosine

7. Two angles whose sum is 90 degrees

8. Value of sine of 30 degrees

9. The value of $(1+ 2 sin 90^0) (1- sin 0^0. sin 10^0. sin 30^0. sin 80^0. sin 90^0)$

1. Ratio of side adjacent to angle and cosine

2. Value of cosine of 90 degrees

4. side opposite to angle

6. Value of cosecant 30 degrees

Check your Answers

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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