Trigonometry for class 10th by Zubair Sir

1. Trigonometry

2. Trigonometry
Tri
Gon
Metron
Three
Sides
Measure
Derived from Three Greek Words

3. Trigonometry
Trigonometry is the branch of mathematics which deals with
triangles particularly triangles in a plane where one angle is
90 degree ( i. e. Right Angled Triangles)
Trigonometry specifically deals with relationships
between the sides and the angles of a triangle.

4. Right- Angled triangle
A
B C
BASE
PERPENDICULAR
Hypotenuse The side opposite to the right angle.
Base The horizontal side of the triangle
Perpendicular The vertical side of the triangle.

5. A
B
BASE
PERPENDICULAR
Right- Angled triangle
A
B C
BASE
PERPENDICULAR

6. Right- Angled triangle
A
B C
BASE
PERPENDICULAR
Ɵ
Hypotenuse The side opposite to the right angle.
Base The side adjacent to active angle.
Perpendicular The side opposite to active angle.
AC
BC
AB

7. Pythagoras Theorem
The square of hypotenuse is equal
to the sum of the squares of other
two sides.
A
B C
𝑨𝑪𝟐 = 𝑩𝑪𝟐 + 𝑨𝑩𝟐

8. Why Trigonometry
In Pythagoras Theorem, we have
given two sides and using
theorem formula, we can find
the third side.
A
B C
But in few situations, we know
the dimension of one side only
and we find other two sides by
Trigonometry.

9. C
B A
Ɵ
Trigonometric Ratios
Trigonometric Ratios of an acute
angle in a right triangle is the
relation between the angle and
the length of its sides.

10. C
B A
Ɵ
Trigonometric Ratios of <A
The trigonometric Ratios of <A in right triangle ABC are defined as follows:
1 =
𝑺𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 < 𝑨
𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝑺𝒊𝒏𝒆 𝒐𝒓 𝑺𝒊𝒏 𝒐𝒇 < 𝑨
𝑶𝒓 𝑺𝒊𝒏𝑨
2 𝑺𝒊𝒅𝒆 𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 < 𝑨
𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
Cos𝒊𝒏𝒆 𝒐𝒓 𝑪𝒐𝒔 𝒐𝒇 < 𝑨
𝑶𝒓 𝑪𝒐𝒔𝑨
3 𝑺𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 < 𝑨
𝑺𝒊𝒅𝒆 𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 < 𝑨
𝑻𝒂𝒏𝒈𝒆𝒏𝒕 𝒐𝒓 𝑻𝒂𝒏 𝒐𝒇 < 𝑨
𝑶𝒓 𝑻𝒂𝒏𝑨
𝑩𝑪
𝑨𝑪
=
=
𝑨𝑩
𝑨𝑪
=
=
𝑩𝑪
𝑨𝑩
=

11. C
B A
Ɵ
Trigonometric Ratios of <A
The trigonometric Ratios of <A in right triangle ABC are defined as follows:
4 =
𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝑺𝒊𝒅𝒆 𝑶𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 < 𝑨
𝑪𝒐𝒔𝒆𝒄𝒂𝒏𝒕 𝒐𝒓 𝑪𝒐𝒔𝒆𝒄 𝒐𝒇 < 𝑨
𝑶𝒓 𝑪𝒐𝒔𝒆𝒄𝑨
5 𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝑺𝒊𝒅𝒆 𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 < 𝑨
𝑺𝒆𝒄𝒂𝒏𝒕 𝒐𝒓 𝑺𝒆𝒄 𝒐𝒇 < 𝑨
𝑶𝒓 𝑺𝒆𝒄𝑨
6 𝑺𝒊𝒅𝒆 𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕𝒕𝒐 < 𝑨
𝑺𝒊𝒅𝒆 𝑶𝒑𝒑𝒐𝒔𝒊𝒕𝒆𝒕𝒐 < 𝑨
Cot𝒂𝒏𝒈𝒆𝒏𝒕 𝒐𝒓 𝑪𝒐𝒕 𝒐𝒇 < 𝑨
𝑶𝒓 𝑪𝒐𝒕𝑨
𝑨𝑪
𝑩𝑪
=
=
𝑨𝑪
𝑨𝑩
=
=
𝑨𝑩
𝑩𝑪
=

12. RECIPROCAL RELATIONS
C
B A
Ɵ
=
𝑺𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 < 𝑨
𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝑺𝒊𝒏 𝑨
𝑩𝑪
𝑨𝑪
=
=
𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝑺𝒊𝒅𝒆 𝑶𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 < 𝑨
𝑪𝒐𝒔𝒆𝒄 𝑨
𝑨𝑪
𝑩𝑪
=
𝑺𝒊𝒏 𝑨
𝟏
𝑪𝒐𝒔𝒆𝒄𝑨
= 𝑪𝒐𝒔𝒆𝒄 𝑨
𝟏
𝑺𝒊𝒏𝑨
=

13. RECIPROCAL RELATIONS
=
𝑺𝒊𝒅𝒆 𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 < 𝑨
𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝑪𝒐𝒔 𝑨
𝑨𝑩
𝑨𝑪
=
=
𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
𝑺𝒊𝒅𝒆 𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 < 𝑨
𝑺𝒆𝒄 𝑨
𝑨𝑪
𝑩𝑪
=
𝑪𝒐𝒔 𝑨
𝟏
𝑺𝒆𝒄𝑨
= 𝑺𝒆𝒄 𝑨
𝟏
𝑪𝒐𝒔𝑨
=
C
B A
Ɵ

14. RECIPROCAL RELATIONS
=
𝑺𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 < 𝑨
𝑺𝒊𝒅𝒆 𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 < 𝑨
𝑻𝒂𝒏𝑨
𝑩𝑪
𝑨𝑩
=
=
𝑺𝒊𝒅𝒆 𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 < 𝑨
𝑺𝒊𝒅𝒆 𝑶𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 < 𝑨
𝑪𝒐𝒕𝑨
𝑨𝑩
𝑩𝑪
=
𝑻𝒂𝒏𝑨
𝟏
𝑪𝒐𝒕𝑨
= C𝒐𝒕 𝑨
𝟏
𝑻𝒂𝒏𝑨
=
C
B A
Ɵ

15. A
B
QUOTIENT RELATIONS
𝑺𝒊𝒏 𝑨
𝑩𝑪
𝑨𝑪
= 𝑪𝒐𝒔 𝑨
𝑨𝑩
𝑨𝑪
=
𝑺𝒊𝒏 𝑨
𝑪𝒐𝒔 𝑨
=
𝑩𝑪
𝑨𝑪
𝑨𝑩
𝑨𝑪
𝑻𝒂𝒏 𝑨
𝑩𝑪
𝑨𝑩
=
=
𝑩𝑪
𝑨𝑩
= 𝑻𝒂𝒏 𝑨
𝑻𝒂𝒏 𝑨 =
𝑺𝒊𝒏 𝑨
𝑪𝒐𝒔 𝑨
C𝒐𝒕 𝑨
𝟏
𝑻𝒂𝒏𝑨
= =
𝑪𝒐𝒔 𝑨
𝑺𝒊𝒏 𝑨

16. Trigonometric ratios of some specific angles
For 𝟒𝟓𝒐
A
B
a
a
𝟒𝟓𝒐
𝟒𝟓𝒐
𝟗𝟎𝒐
Let ABC be an isosceles triangle right angled at B.
Then, <A = 𝟒𝟓𝒐
and <C = 𝟒𝟓𝒐
Applying Pythagoras Theorem to ∆𝑨𝑩𝑪,We get
𝑨𝑪𝟐 = 𝑨𝑩𝟐 + 𝑩𝑪𝟐
⇒ 𝑨𝑪𝟐
= 𝒂𝟐
+ 𝒂𝟐
⇒ 𝑨𝑪𝟐= 𝟐𝒂𝟐
⇒ 𝑨𝑪 = 𝟐𝒂
⇒ 𝑨𝑪 = 𝟐𝒂𝟐

17. Trigonometric ratios of some specific angles
For 𝟒𝟓𝒐
A B
a
a
𝟒𝟓𝒐
𝟒𝟓𝒐
𝟗𝟎𝒐
𝑺𝒊𝒏 𝟒𝟓𝒐 𝑩𝑪
𝑨𝑪
=
𝑪𝒐𝒔 𝟒𝟓𝒐 𝑨𝑩
𝑨𝑪
=
𝑻𝒂𝒏 𝟒𝟓𝒐 𝑩𝑪
𝑨𝑩
=
Cosec 𝟒𝟓𝒐 𝑨𝑪
𝑩𝑪
=
𝑺𝒆𝒄 𝟒𝟓𝒐 𝑨𝑪
𝑨𝑩
=
𝑪𝒐𝒕 𝟒𝟓𝒐 𝑨𝑩
𝑩𝑪
=
𝒂
𝟐𝒂
=
𝒂
𝟐𝒂
=
𝒂
𝒂
=
𝒂
𝒂
=
=
𝟐𝒂
𝒂
=
𝟐𝒂
𝒂
𝑻𝒂𝒏 𝟒𝟓𝒐
= 1 𝑪𝒐𝒕 𝟒𝟓𝒐
= 1
𝑪𝒐𝒔 𝟒𝟓𝒐
=
𝟏
𝟐
𝑺𝒆𝒄 𝟒𝟓𝒐
=
𝑺𝒊𝒏 𝟒𝟓𝒐
=
𝟏
𝟐
𝑪𝒐𝒔𝒆𝒄 𝟒𝟓𝒐
= 𝟐
𝟐

18. Trigonometric ratios of some specific angles
For 𝟔𝟎𝒐
𝒂𝒏𝒅 𝟑𝟎𝒐
Let ABC be an equilateral triangle.
Then, <A = 𝟔𝟎𝒐
, <B = 𝟔𝟎𝒐
and <C = 𝟔𝟎𝒐
Let each side of the triangle is of length 2a
A B
𝟔𝟎𝒐
𝟔𝟎𝒐
𝟑𝟎𝒐
𝟐𝒂
𝟐𝒂
𝟐𝒂
Now draw an altitude CD from vertex C to opposite side AB
D
𝒂 𝒂
𝟑𝟎𝒐
Applying Pythagoras Theorem to ∆𝑪𝑫𝑩,We get
𝑩𝑪𝟐 = 𝑩𝑫𝟐 + 𝑫𝑪𝟐
⇒ (𝟐𝒂)𝟐= 𝒂𝟐 + 𝑫𝑪𝟐
⇒ 𝑫𝑪 = 𝟑𝒂𝟐
⇒ 𝟒𝒂𝟐 − 𝒂𝟐 = 𝑫𝑪𝟐
⇒ 𝟑𝒂𝟐= 𝑫𝑪𝟐
⇒ 𝑫𝑪 = 𝟑𝒂
𝟑𝒂

19. Trigonometric ratios of some specific angles
For 𝟔𝟎𝒐
𝟔𝟎𝒐
𝟐𝒂
D 𝒂
𝟑𝟎𝒐
𝟑𝒂
B
C
𝑺𝒊𝒏 𝟔𝟎𝒐 𝑫𝑪
𝑩𝑪
=
𝑪𝒐𝒔 𝟔𝟎𝒐 𝑩𝑫
𝑩𝑪
=
𝑻𝒂𝒏 𝟔𝟎𝒐 𝑫𝑪
𝑩𝑫
=
Cosec 𝟔𝟎𝒐 𝑩𝑪
𝑫𝑪
=
𝑺𝒆𝒄 𝟔𝟎𝒐 𝑩𝑪
𝑩𝑫
=
𝑪𝒐𝒕 𝟔𝟎𝒐 𝑩𝑫
𝑫𝑪
=
𝟑𝒂
𝟐𝒂
=
𝒂
𝟐𝒂
=
𝟑𝒂
𝒂
=
𝒂
𝟑𝒂
=
=
𝟐𝒂
𝟑𝒂
=
𝟐𝒂
𝒂
𝑻𝒂𝒏 𝟔𝟎𝒐
= 𝟑
𝑪𝒐𝒔 𝟔𝟎𝒐
= 𝟏
𝟐
𝑺𝒆𝒄 𝟔𝟎𝒐
=
𝑺𝒊𝒏 𝟔𝟎𝒐
= 𝟑
𝟐
𝑪𝒐𝒔𝒆𝒄 𝟔𝟎𝒐
=
𝟐
𝟑
𝟐
𝑪𝒐𝒕 𝟔𝟎𝒐
=
𝟏
𝟑

20. Trigonometric ratios of some specific angles
For 𝟑𝟎𝒐
𝟔𝟎𝒐
𝟐𝒂
D 𝒂
𝟑𝟎𝒐
𝟑𝒂
B
C
𝑺𝒊𝒏 𝟑𝟎𝒐 𝑩𝑫
𝑩𝑪
=
𝑪𝒐𝒔 𝟑𝟎𝒐 𝑫𝑪
𝑩𝑪
=
𝑻𝒂𝒏 𝟑𝟎𝒐 𝑩𝑫
𝑫𝑪
=
Cosec 𝟑𝟎𝒐 𝑩𝑪
𝑩𝑫
=
𝑺𝒆𝒄 𝟑𝟎𝒐 𝑩𝑪
𝑫𝑪
=
𝑪𝒐𝒕 𝟑𝟎𝒐 𝑫𝑪
𝑩𝑫
=
𝒂
𝟐𝒂
=
𝟑𝒂
𝟐𝒂
=
𝒂
𝟑𝒂
=
𝟑𝒂
𝒂
=
=
𝟐𝒂
𝒂
=
𝟐𝒂
𝟑𝒂
𝑻𝒂𝒏 𝟑𝟎𝒐
=
𝟏
𝟑
𝑪𝒐𝒔 𝟑𝟎𝒐
= 𝟑
𝟐
𝑺𝒆𝒄 𝟑𝟎𝒐
=
𝑺𝒊𝒏 𝟑𝟎𝒐
= 𝟏
𝟐
𝑪𝒐𝒔𝒆𝒄 𝟑𝟎𝒐
=
𝟐
𝟑
𝑪𝒐𝒕 𝟑𝟎𝒐
=
𝟐
𝟑

21. Trigonometric ratios of some specific angles
For 𝟎𝒐
A B
Ɵ𝒐
In ∆𝑨𝑩𝑪, <A will be 𝟎𝒐 when AC and AB coincides,
That is AC = AB
A B
And BC will be of length zero. I.e. BC = 0

22. Trigonometric ratios of some specific angles
For 𝟎𝒐
A B
𝟎𝒐
𝑺𝒊𝒏 𝟎𝒐 𝑩𝑪
𝑨𝑪
=
𝑪𝒐𝒔 𝟎𝒐 𝑨𝑩
𝑨𝑪
𝑻𝒂𝒏 𝟎𝒐 𝑩𝑪
𝑨𝑩
=
Cosec 𝟎𝒐 𝑨𝑪
𝑩𝑪
=
𝑺𝒆𝒄 𝟎𝒐 𝑨𝑪
𝑨𝑩
=
𝑪𝒐𝒕 𝟎𝒐 𝑨𝑩
𝑩𝑪
=
𝟎
𝑨𝑪
=
𝟎
𝑨𝑩
=
𝑨𝑩
𝟎
=
𝑨𝑪
𝟎
=
𝑻𝒂𝒏 𝟎𝒐
= 0 𝑪𝒐𝒕 𝟎𝒐
=
𝑪𝒐𝒔 𝟎𝒐
= 𝟏 𝑺𝒆𝒄 𝟎𝒐
=
𝑺𝒊𝒏 𝟎𝒐
= 𝟎 𝑪𝒐𝒔𝒆𝒄 𝟎𝒐
= 𝑵𝒐𝒕 𝑫𝒆𝒇𝒊𝒏𝒆𝒅
𝟏
=
𝑵𝒐𝒕 𝑫𝒆𝒇𝒊𝒏𝒆𝒅

23. Trigonometric ratios of some specific angles
For 𝟗𝟎𝒐
A B
Ɵ𝒐
In ∆𝑨𝑩𝑪, <A will be 𝟗𝟎𝒐 when AC and BC coincides,
That is AC = BC
A B
And AB will be of length zero. I.e. BA = 0

24. Trigonometric ratios of some specific angles
For 𝟗𝟎𝒐
A
B
𝑺𝒊𝒏 𝟗𝟎𝒐 𝑩𝑪
𝑨𝑪
=
𝑪𝒐𝒔 𝟗𝟎𝒐 𝑨𝑩
𝑨𝑪
𝑻𝒂𝒏 𝟗𝟎𝒐 𝑩𝑪
𝑨𝑩
=
Cosec 𝟗𝟎𝒐 𝑨𝑪
𝑩𝑪
=
𝑺𝒆𝒄 𝟗𝟎𝒐 𝑨𝑪
𝑨𝑩
=
𝑪𝒐𝒕 𝟗𝟎𝒐 𝑨𝑩
𝑩𝑪
=
𝟎
𝑨𝑪
=
𝑩𝑪
𝟎
=
𝟎
𝑩𝑪
=
𝑨𝑪
𝟎
=
𝑻𝒂𝒏 𝟗𝟎𝒐
= 0
𝑪𝒐𝒕 𝟗𝟎𝒐
=
𝑪𝒐𝒔 𝟗𝟎𝒐
= 𝟎 𝑺𝒆𝒄 𝟗𝟎𝒐
=
𝑺𝒊𝒏 𝟗𝟎𝒐
= 𝟏 𝑪𝒐𝒔𝒆𝒄 𝟗𝟎𝒐
=
𝑵𝒐𝒕 𝑫𝒆𝒇𝒊𝒏𝒆𝒅
𝟏
=
𝑵𝒐𝒕 𝑫𝒆𝒇𝒊𝒏𝒆𝒅

25. Angle
T-Ratios 𝟎𝒐
𝟑𝟎𝒐
𝟒𝟓𝒐
𝟔𝟎𝒐
𝟗𝟎𝒐
Sin Ɵ 0 𝟏
𝟐
𝟏
𝟐
𝟑
𝟐
1
Cos Ɵ 1 𝟑
𝟐
𝟏
𝟐
𝟏
𝟐
0
Tan Ɵ 0 𝟏
𝟑
1 𝟑 Not Defined
Cosec Ɵ Not Defined 2 𝟐 𝟐
𝟑
1
Sec Ɵ 1 𝟐
𝟑
𝟐 2 Not Defined
Cot Ɵ Not Defined 𝟑 1 𝟏
𝟑
0

26. Trigonometric ratios of Complimentary Angles
A B
∆𝑨𝑩𝑪 𝐢𝐬 𝐚 𝐫𝐢𝐠𝐡𝐭 𝐚𝐧𝐠𝐥𝐞𝐝 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐰𝐢𝐭𝐡 <B =𝟗𝟎𝒐
Now, in ∆𝑨𝑩𝑪 , 𝐔𝐬𝐢𝐧𝐠 𝐀𝐧𝐠𝐥𝐞 𝐒𝐮𝐦 𝐏𝐫𝐨𝐩𝐞𝐫𝐭𝐲.
< 𝑨+ < 𝑩+ < 𝑪 = 𝟏𝟖𝟎𝒐
Ɵ + 𝟗𝟎𝒐
+ < 𝑪 = 𝟏𝟖𝟎𝒐
Ɵ+ < 𝑪 = 𝟏𝟖𝟎𝒐
− 𝟗𝟎𝒐
< 𝑪 = 𝟗𝟎𝒐 − Ɵ
𝟗𝟎𝒐
− Ɵ
Ɵ
Ɵ+ < 𝑪 = 𝟗𝟎𝒐

27. Trigonometric ratios of Complimentary Angles
A B
𝟗𝟎𝒐
− Ɵ
𝑺𝒊𝒏Ɵ
𝑩𝑪
𝑨𝑪
= 𝑪𝒐𝒔Ɵ
𝑨𝑩
𝑨𝑪
𝑻𝒂𝒏Ɵ
𝑩𝑪
𝑨𝑩
=
CosecƟ
𝑨𝑪
𝑩𝑪
= 𝑺𝒆𝒄Ɵ
𝑨𝑪
𝑨𝑩
= 𝑪𝒐𝒕Ɵ
𝑨𝑩
𝑩𝑪
=
=
𝑺𝒊𝒏(𝟗𝟎 − Ɵ)
𝑩𝑪
𝑨𝑪
=
𝑪𝒐𝒔(𝟗𝟎 − Ɵ)
𝑨𝑩
𝑨𝑪
𝑻𝒂𝒏Ɵ
𝑩𝑪
𝑨𝑩
CosecƟ
𝑨𝑪
𝑩𝑪
=
𝑺𝒆𝒄Ɵ
𝑨𝑪
𝑨𝑩
= 𝑪𝒐𝒕Ɵ
𝑨𝑩
𝑩𝑪
=
=
𝑪𝒐𝒔Ɵ
= =
𝑺𝒊𝒏Ɵ
𝑻𝒂𝒏 (𝟗𝟎 − Ɵ)
𝑪𝒐𝒔𝒆𝒄(𝟗𝟎 − Ɵ)
Ɵ
=
=
= S𝒆𝒄(𝟗𝟎 − Ɵ) =
𝑪𝒐𝒕(𝟗𝟎 − Ɵ) =

28. Trigonometric ratios of Complimentary Angles
A B
𝟗𝟎𝒐
− Ɵ
𝑺𝒊𝒏(𝟗𝟎 − Ɵ)
𝑪𝒐𝒔(𝟗𝟎 − Ɵ)
𝑻𝒂𝒏Ɵ
CosecƟ
=
𝑺𝒆𝒄Ɵ
𝑪𝒐𝒕Ɵ
=
𝑪𝒐𝒔Ɵ
=
=
𝑺𝒊𝒏Ɵ
𝑻𝒂𝒏 (𝟗𝟎 − Ɵ)
𝑪𝒐𝒔𝒆𝒄(𝟗𝟎 − Ɵ)
Ɵ
=
=
S𝒆𝒄(𝟗𝟎 − Ɵ)
𝑪𝒐𝒕(𝟗𝟎 − Ɵ)

29. Trigonometric ratios of Complimentary Angles
A B
𝟗𝟎𝒐
− Ɵ
𝑺𝒊𝒏(𝟗𝟎 − Ɵ)
𝑪𝒐𝒔(𝟗𝟎 − Ɵ)
𝑻𝒂𝒏Ɵ
CosecƟ
=
𝑺𝒆𝒄Ɵ
𝑪𝒐𝒕Ɵ
=
𝑪𝒐𝒔Ɵ
=
=
𝑺𝒊𝒏Ɵ
𝑻𝒂𝒏 (𝟗𝟎 − Ɵ)
𝑪𝒐𝒔𝒆𝒄(𝟗𝟎 − Ɵ)
Ɵ
=
=
S𝒆𝒄(𝟗𝟎 − Ɵ)
𝑪𝒐𝒕(𝟗𝟎 − Ɵ)
Questions
𝟏. 𝑰𝒇 𝑻𝒂𝒏 𝑨 = 𝑪𝒐𝒕 𝑩, 𝒕𝒉𝒆𝒏 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝑨 + 𝑩

30. Trigonometric ratios of Complimentary Angles
A B
𝟗𝟎𝒐
− Ɵ
𝑺𝒊𝒏(𝟗𝟎 − Ɵ)
𝑪𝒐𝒔(𝟗𝟎 − Ɵ)
𝑻𝒂𝒏Ɵ
CosecƟ
=
𝑺𝒆𝒄Ɵ
𝑪𝒐𝒕Ɵ
=
𝑪𝒐𝒔Ɵ
=
=
𝑺𝒊𝒏Ɵ
𝑻𝒂𝒏 (𝟗𝟎 − Ɵ)
𝑪𝒐𝒔𝒆𝒄(𝟗𝟎 − Ɵ)
Ɵ
=
=
S𝒆𝒄(𝟗𝟎 − Ɵ)
𝑪𝒐𝒕(𝟗𝟎 − Ɵ)
Questions
2. 𝑰𝒇 𝑻𝒂𝒏 𝟐𝑨 = 𝑪𝒐𝒕 (𝑨 − 𝟏𝟖𝒐), 𝒕𝒉𝒆𝒏 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝑨

31. A B
Ɵ
TRIGONOMETRIC IDENTITIES
∆𝑨𝑩𝑪 𝐢𝐬 𝐚 𝐫𝐢𝐠𝐡𝐭 𝐚𝐧𝐠𝐥𝐞𝐝 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐰𝐢𝐭𝐡 <B =𝟗𝟎𝒐
Now, in ∆𝑨𝑩𝑪 , 𝐔𝐬𝐢𝐧𝐠 𝐏𝐲𝐭𝐡𝐚𝐠𝐨𝐫𝐮𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦.
𝑨𝑩𝟐
+ 𝑩𝑪𝟐
= 𝑨𝑪𝟐
−− −(𝒊)
𝑺𝒊𝒏Ɵ
𝑩𝑪
𝑨𝑪
= 𝑪𝒐𝒔Ɵ
𝑨𝑩
𝑨𝑪
𝑻𝒂𝒏Ɵ
𝑩𝑪
𝑨𝑩
= CosecƟ
𝑨𝑪
𝑩𝑪
= 𝑺𝒆𝒄Ɵ
𝑨𝑪
𝑨𝑩
= 𝑪𝒐𝒕Ɵ
𝑨𝑩
𝑩𝑪
=
=
Divide (i) by 𝑨𝑩𝟐
, we get:
𝑨𝑩𝟐
𝑨𝑩𝟐
+
𝑩𝑪𝟐
𝑨𝑩𝟐
=
𝑨𝑪𝟐
𝑨𝑩𝟐
𝟏 +
𝑩𝑪
𝑨𝑩
𝟐
=
𝑨𝑪
𝑨𝑩
𝟐
𝟏 + 𝒕𝒂𝒏𝟐
Ɵ = 𝒔𝒆𝒄𝟐
Ɵ

32. A B
Ɵ
TRIGONOMETRIC IDENTITIES
∆𝑨𝑩𝑪 𝐢𝐬 𝐚 𝐫𝐢𝐠𝐡𝐭 𝐚𝐧𝐠𝐥𝐞𝐝 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐰𝐢𝐭𝐡 <B =𝟗𝟎𝒐
Now, in ∆𝑨𝑩𝑪 , 𝐔𝐬𝐢𝐧𝐠 𝐏𝐲𝐭𝐡𝐚𝐠𝐨𝐫𝐮𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦.
𝑨𝑩𝟐
+ 𝑩𝑪𝟐
= 𝑨𝑪𝟐
−− −(𝒊)
𝑺𝒊𝒏Ɵ
𝑩𝑪
𝑨𝑪
= 𝑪𝒐𝒔Ɵ
𝑨𝑩
𝑨𝑪
𝑻𝒂𝒏Ɵ
𝑩𝑪
𝑨𝑩
= CosecƟ
𝑨𝑪
𝑩𝑪
= 𝑺𝒆𝒄Ɵ
𝑨𝑪
𝑨𝑩
= 𝑪𝒐𝒕Ɵ
𝑨𝑩
𝑩𝑪
=
=
Divide (i) by 𝑩𝑪𝟐
, we get:
𝑨𝑩𝟐
𝑩𝑪𝟐
+
𝑩𝑪𝟐
𝑩𝑪𝟐
=
𝑨𝑪𝟐
𝑩𝑪𝟐
𝑨𝑩
𝑩𝑪
𝟐
+ 𝟏 =
𝑨𝑪
𝑩𝑪
𝟐
𝒄𝒐𝒕𝟐
Ɵ + 𝟏 = 𝒄𝒐𝒔𝒆𝒄𝟐
Ɵ

33. A B
Ɵ
TRIGONOMETRIC IDENTITIES
∆𝑨𝑩𝑪 𝐢𝐬 𝐚 𝐫𝐢𝐠𝐡𝐭 𝐚𝐧𝐠𝐥𝐞𝐝 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐰𝐢𝐭𝐡 <B =𝟗𝟎𝒐
Now, in ∆𝑨𝑩𝑪 , 𝐔𝐬𝐢𝐧𝐠 𝐏𝐲𝐭𝐡𝐚𝐠𝐨𝐫𝐮𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦.
𝑨𝑩𝟐
+ 𝑩𝑪𝟐
= 𝑨𝑪𝟐
−− −(𝒊)
𝑺𝒊𝒏Ɵ
𝑩𝑪
𝑨𝑪
= 𝑪𝒐𝒔Ɵ
𝑨𝑩
𝑨𝑪
𝑻𝒂𝒏Ɵ
𝑩𝑪
𝑨𝑩
= CosecƟ
𝑨𝑪
𝑩𝑪
= 𝑺𝒆𝒄Ɵ
𝑨𝑪
𝑨𝑩
= 𝑪𝒐𝒕Ɵ
𝑨𝑩
𝑩𝑪
=
=
Divide (i) by 𝑨𝑪𝟐
, we get:
𝑨𝑩𝟐
𝑨𝑪𝟐
+
𝑩𝑪𝟐
𝑨𝑪𝟐
=
𝑨𝑪𝟐
𝑨𝑪𝟐
𝑨𝑩
𝑨𝑪
𝟐
+
𝑩𝑪
𝑨𝑪
𝟐
= 𝟏
𝒔𝒊𝒏𝟐
Ɵ + 𝒄𝒐𝒔𝟐
Ɵ = 𝟏

34. Trigonometric ratios of Complimentary Angles
A B
𝟗𝟎𝒐
− Ɵ
𝑺𝒊𝒏(𝟗𝟎 − Ɵ)
𝑪𝒐𝒔(𝟗𝟎 − Ɵ)
𝑻𝒂𝒏Ɵ
CosecƟ
=
𝑺𝒆𝒄Ɵ
𝑪𝒐𝒕Ɵ
=
𝑪𝒐𝒔Ɵ
=
=
𝑺𝒊𝒏Ɵ
𝑻𝒂𝒏 (𝟗𝟎 − Ɵ)
𝑪𝒐𝒔𝒆𝒄(𝟗𝟎 − Ɵ)
Ɵ
=
=
S𝒆𝒄(𝟗𝟎 − Ɵ)
𝑪𝒐𝒕(𝟗𝟎 − Ɵ)
Questions
𝟏. 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 (𝟏 + 𝒕𝒂𝒏Ɵ + 𝐬𝐞𝐜Ɵ )(𝟏 + 𝒄𝒐𝒕Ɵ − 𝒄𝒐𝒔𝒆𝒄Ɵ)

35. Trigonometric ratios of Complimentary Angles
A B
𝟗𝟎𝒐
− Ɵ
𝑺𝒊𝒏(𝟗𝟎 − Ɵ)
𝑪𝒐𝒔(𝟗𝟎 − Ɵ)
𝑻𝒂𝒏Ɵ
CosecƟ
=
𝑺𝒆𝒄Ɵ
𝑪𝒐𝒕Ɵ
=
𝑪𝒐𝒔Ɵ
=
=
𝑺𝒊𝒏Ɵ
𝑻𝒂𝒏 (𝟗𝟎 − Ɵ)
𝑪𝒐𝒔𝒆𝒄(𝟗𝟎 − Ɵ)
Ɵ
=
=
S𝒆𝒄(𝟗𝟎 − Ɵ)
𝑪𝒐𝒕(𝟗𝟎 − Ɵ)
Questions
𝟐. 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 (𝐒𝐞𝐜𝐀 + 𝐭𝐚𝐧𝐀)(𝟏 − 𝐬𝐢𝐧𝐀)

36. Trigonometric ratios of Complimentary Angles
A B
𝟗𝟎𝒐
− Ɵ
𝑺𝒊𝒏(𝟗𝟎 − Ɵ)
𝑪𝒐𝒔(𝟗𝟎 − Ɵ)
𝑻𝒂𝒏Ɵ
CosecƟ
=
𝑺𝒆𝒄Ɵ
𝑪𝒐𝒕Ɵ
=
𝑪𝒐𝒔Ɵ
=
=
𝑺𝒊𝒏Ɵ
𝑻𝒂𝒏 (𝟗𝟎 − Ɵ)
𝑪𝒐𝒔𝒆𝒄(𝟗𝟎 − Ɵ)
Ɵ
=
=
S𝒆𝒄(𝟗𝟎 − Ɵ)
𝑪𝒐𝒕(𝟗𝟎 − Ɵ)
Questions
𝟑. 𝐖𝐫𝐢𝐭𝐞 𝐚𝐥𝐥 𝐨𝐭𝐡𝐞𝐫 𝐭𝐢𝐠𝐨𝐧𝐨𝐦𝐞𝐭𝐫𝐢𝐜 𝐫𝐚𝐭𝐢𝐨𝐬 𝐨𝐟 < 𝐀 𝐢𝐧 𝐭𝐞𝐫𝐦𝐬 𝐨𝐟 𝐒𝐞𝐜𝐀.

37. Thank You